NCERT Class 12 Maths Integrals Ex 7.5 Solution

📘 Concept & Theory

Theory and Concept

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Integration of Rational Functions by Partial Fractions

A rational function is a function expressed as the ratio of two polynomials. When the degree of the numerator is less than the degree of the denominator, the rational function is called a proper rational function.

Some rational functions cannot be integrated directly using standard integration formulae. In such cases, we simplify the integrand by expressing it as a sum of simpler fractions. This technique is known as partial fraction decomposition.

For two distinct linear factors, the standard decomposition is:

\[ \frac{P(x)}{(x+a)(x+b)} = \frac{A}{x+a}+\frac{B}{x+b} \]

Here, A and B are constants determined by comparing coefficients or substituting suitable values of x.

Important integration formula:

\[ \int \frac{1}{x+a}\,dx = \log|x+a|+C \]

Thus, once a rational function is decomposed into simpler fractions with linear denominators, each term can be integrated using the logarithmic integration formula.

Conditions to remember:

The degree of the numerator should be less than the degree of the denominator for direct partial fraction decomposition.
Distinct linear factors have separate constant numerators.
The constants are determined by comparing coefficients or substituting convenient values.
The absolute value sign must be included in logarithmic expressions to account for the domain of the logarithm.
The constant of integration C must be included in the final answer.

🗺️ Solution Roadmap

Step-by-step Plan

›

Identify the rational function and observe that its denominator contains two distinct linear factors.
Express the rational function as the sum of two partial fractions with unknown constants A and B.
Multiply both sides by the common denominator to eliminate fractions.
Compare the coefficients of x and the constant terms to obtain two simultaneous linear equations.
Solve the equations to determine the values of A and B.
Substitute the values into the original integral and split it into two simpler integrals.
Apply the standard logarithmic integration formula and simplify the result.
Include the absolute value signs and the constant of integration.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 21 steps

Given —
\[I=\int \frac{x}{(x+1)(x+2)}\,dx\]
Apply the method of partial fractions.
Since the denominator consists of two distinct linear factors, we assume:
\[\frac{x}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}\]
Taking the LCM on the right-hand side, we get:
\[\frac{x}{(x+1)(x+2)}=\frac{A(x+2)+B(x+1)}{(x+1)(x+2)}\]
Since the denominators on both sides are identical, equating the numerators gives:
\[x=A(x+2)+B(x+1)\]
Expand and collect like terms.
\[x=Ax+2A+Bx+B\]
Rearranging and collecting the coefficients of x and the constant terms:
\[x=(A+B)x+(2A+B)\]
Compare coefficients
Comparing the coefficients of x on both sides:
\[A+B=1 \tag{1}\]
Comparing the constant terms:
\[2A+B=0 \tag{2}\]
Solve the simultaneous equations
From equation (1):
\[A+B=1\]

\[B=1-A \tag{3}\]
Substituting equation (3) into equation (2):
\[2A+B=0\]

\[2A+(1-A)=0\]
Removing the brackets:
\[2A+1-A=0\]

\[A+1=0\]
Therefore:
\[\boxed{A=-1}\]
Substituting A = -1 into equation (3):
\[B=1-(-1)\]

\[\boxed{B=2}\]
Hence, the required partial fraction decomposition is:
\[\boxed{\frac{x}{(x+1)(x+2)}=-\frac{1}{x+1}+\frac{2}{x+2}}\]
Substitute the partial fractions into the integral
We have:
\[I=\int \frac{x}{(x+1)(x+2)}\,dx\]
Using the partial fraction decomposition:
\[I=\int\left(-\frac{1}{x+1}+\frac{2}{x+2}\right)dx\]
Applying the linearity property of integration:
\[I=-\int\frac{1}{x+1}\,dx+2\int\frac{1}{x+2}\,dx\]
Apply the standard logarithmic integration formula.
\[\int\frac{1}{x+a}\,dx=\log|x+a|+C\]
Therefore, integrating both terms separately:
\[I=-\log|x+1|+2\log|x+2|+C\]
Simplify the logarithmic expression
Using the logarithmic identity:
\[n\log a=\log(a^n)\]
We obtain:
\[I=-\log|x+1|+\log|x+2|^2+C\]
Using the property:
\[\log a-\log b=\log\left(\frac{a}{b}\right)\]
we get:
\[I=\log\left(\frac{|x+2|^2}{|x+1|}\right)+C\]
Since |x + 2|² = |(x + 2)²|, the final result can be written as:
\[\boxed{\displaystyle\int\frac{x}{(x+1)(x+2)}\,dx=\log\left|\frac{(x+2)^2}{x+1}\right|+C}\]

💡 Answer

Final Answer

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\[ \boxed{ \displaystyle \int\frac{x}{(x+1)(x+2)}\,dx = \log\left| \frac{(x+2)^2}{x+1} \right|+C } \]

🎯 Exam Significance

Exam Significance

›

Application of integration techniques: It tests the ability to recognise when a rational function should be decomposed into partial fractions.
Algebraic accuracy: Students must correctly expand expressions, collect like terms and solve simultaneous linear equations.
Logarithmic integration: It reinforces the standard integral of 1/(x + a) and the properties of logarithms.
Stepwise presentation: Writing the decomposition, determining constants and integrating each term separately makes the solution logically complete and suitable for examination evaluation.
Verification skills: Differentiating the final answer provides an effective way to check the correctness of an indefinite integral.

Significance for Competitive Entrance Examinations

The concepts used in this question are relevant to JEE Main, JEE Advanced, BITSAT and other engineering entrance examinations.

Partial fraction decomposition: Questions involving rational functions frequently require the identification of suitable partial fraction forms.
Algebraic manipulation: Efficient determination of unknown constants helps reduce calculation time in objective-type questions.
Integration and differentiation connection: The ability to verify an antiderivative by differentiation is useful in solving integral-based problems.
Logarithmic identities: Combining logarithmic terms correctly is essential in simplifying final expressions and comparing equivalent answer choices.
Foundation for advanced problems: The same decomposition technique is useful in more complicated rational integrals involving repeated linear factors and irreducible quadratic factors.

🚫 Common Mistakes

Common Mistakes

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Incorrect partial fraction assumption: For distinct linear factors, use separate constant numerators over each factor.
Sign errors: While expanding A(x + 2) + B(x + 1), ensure that both constant terms are included correctly.
Incorrect coefficient comparison: Remember that coefficients of x and constant terms must be compared separately.
Missing integration constants: The final indefinite integral must contain only one arbitrary constant C.
Ignoring absolute values: Write log|x + 1| and log|x + 2| rather than omitting the absolute value signs.
Incorrect logarithmic simplification: The coefficient 2 in front of log|x + 2| becomes the exponent 2 inside the logarithm.
Incomplete verification: Always ensure that differentiating the final expression reproduces the original integrand.

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 9 points

Partial fraction decomposition converts a rational function into simpler fractions that can be integrated separately.
A rational function with distinct linear factors in its denominator can be decomposed into fractions with constant numerators.
The constants in a partial fraction decomposition can be determined by comparing coefficients or substituting suitable values of the variable.
The fundamental logarithmic integration formula is:
\[ \boxed{ \int\frac{1}{x+a}\,dx=\log|x+a|+C } \]
Logarithmic identities help combine the integrated terms into a compact expression.
The absolute value sign is essential when the logarithm involves a variable expression that may be positive or negative.
Differentiation is an effective method for verifying the correctness of an indefinite integral.
Mastering partial fractions provides a foundation for evaluating more complex rational integrals in board and competitive examinations.

📘 Concept & Theory

Theory and Concept

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Integration of Rational Functions Using Partial Fractions

A rational function is an algebraic expression represented as the ratio of two polynomials. When the denominator can be factorised into distinct linear factors, partial fraction decomposition provides an effective method for evaluating its integral.

In this question, the denominator is a difference of two squares. We use the algebraic identity:

\[ a^2-b^2=(a-b)(a+b) \]

Therefore, the denominator can be factorised as:

\[ x^2-9=x^2-3^2=(x-3)(x+3) \]

Since the denominator contains two distinct linear factors, the rational function can be expressed in the form:

\[ \frac{1}{(x-3)(x+3)} = \frac{A}{x-3}+\frac{B}{x+3} \]

Here, A and B are constants that can be determined by comparing coefficients or substituting suitable values of x.

Important integration formula:

\[ \int\frac{1}{x-a}\,dx = \log|x-a|+C \]

Similarly,

\[ \int\frac{1}{x+a}\,dx = \log|x+a|+C \]

These logarithmic integration formulae are fundamental when integrating rational functions containing distinct linear factors.

🗺️ Solution Roadmap

Step-by-step Plan

›

Identify the denominator and factorise it using the difference of two squares identity.
Express the resulting rational function as the sum of two partial fractions with unknown constants A and B.
Multiply both sides by the common denominator to eliminate fractions.
Compare the coefficients of x and the constant terms to obtain two simultaneous equations.
Solve the equations to determine the values of A and B.
Substitute these values into the original integral and separate it into two simpler integrals.
Apply the standard logarithmic integration formula to each term.
Combine the logarithmic expressions using logarithmic identities and write the final answer with the absolute value sign and integration constant.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 30 steps

Given —
\[I=\int\frac{dx}{x^2-9}\]
Factorise the denominator
Using the difference of two squares identity:
\[a^2-b^2=(a-b)(a+b)\]
We have:
\[x^2-9=x^2-3^2\]
Therefore:
\[x^2-9=(x-3)(x+3)\]
Substituting this factorisation into the integral:
\[I=\int\frac{dx}{(x-3)(x+3)}\]
Apply partial fraction decomposition
Since the denominator contains two distinct linear factors, assume:
\[\frac{1}{(x-3)(x+3)}=\frac{A}{x-3}+\frac{B}{x+3}\]
Taking the LCM on the right-hand side:
\[\frac{1}{(x-3)(x+3)}=\frac{A(x+3)+B(x-3)}{(x-3)(x+3)}\]
Since the denominators on both sides are identical, equating the numerators gives:
\[1=A(x+3)+B(x-3)\]
Expand and collect like terms
Expanding the right-hand side:
\[1=Ax+3A+Bx-3B\]
Rearranging and collecting the coefficients of x:
\[1=x(A+B)+3A-3B\]
Compare coefficients
Comparing the coefficients of x on both sides:
\[A+B=0 \tag{1}\]
Comparing the constant terms:
\[3A-3B=1 \tag{2}\]
Determine the values of A and B.
From equation (1):
\[A+B=0\]
Therefore:
\[A=-B \tag{3}\]
Substituting equation (3) into equation (2):
\[3A-3B=1\]
.
\[3(-B)-3B=1\]
Removing the brackets:
\[-3B-3B=1\]
Combining like terms:
\[-6B=1\]
Dividing both sides by -6:
\[\boxed{B=-\frac{1}{6}}\]
Using equation (3):
\[A=-B\]

\[A=-\left(-\frac{1}{6}\right)\]
Hence:
\[\boxed{A=\frac{1}{6}}\]
Thus, the required partial fraction decomposition is:
\[\boxed{\frac{1}{(x-3)(x+3)}=\frac{1}{6(x-3)}-\frac{1}{6(x+3)}}\]
Substitute the partial fractions into the integral
We know that:
\[I=\int\frac{dx}{(x-3)(x+3)}\]
Using the partial fraction decomposition:
\[I=\int\left(\frac{1}{6(x-3)}-\frac{1}{6(x+3)}\right)dx\]
Taking the constant coefficients outside the integrals:
\[I=\frac{1}{6}\int\frac{dx}{x-3}-\frac{1}{6}\int\frac{dx}{x+3}\]
Apply the logarithmic integration formula.
\[\int\frac{1}{x+a}\,dx=\log|x+a|+C\]
Similarly:
\[\int\frac{1}{x-a}\,dx=\log|x-a|+C\]
Applying these formulae to the two integrals:
\[I=\frac{1}{6}\log|x-3|-\frac{1}{6}\log|x+3|+C\]
Taking 1/6 common:
\[I=\frac{1}{6}\left[\log|x-3|-\log|x+3|\right]+C\]
Simplify using logarithmic identities
Using the identity:
\[\log a-\log b=\log\left(\frac{a}{b}\right)\]
We obtain:
\[I=\frac{1}{6}\log\left(\frac{|x-3|}{|x+3|}\right)+C\]
Since the quotient of absolute values can be expressed as the absolute value of the quotient:
\[I=\frac{1}{6}\log\left|\frac{x-3}{x+3}\right|+C\]
Therefore, the final answer is:
\[\boxed{\displaystyle\int\frac{dx}{x^2-9}=\frac{1}{6}\log\left|\frac{x-3}{x+3}\right|+C}\]

💡 Answer

Final Answer

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🎯 Exam Significance

Exam Significance

›

This problem is important for CBSE Class 12 Mathematics board examination preparation because it combines algebraic factorisation, partial fraction decomposition and logarithmic integration.

Factorisation skills: It tests the ability to recognise and apply the difference of two squares identity.
Partial fraction decomposition: Students learn how to express a rational function with distinct linear factors as a sum of simpler fractions.
Simultaneous linear equations: The determination of A and B reinforces algebraic manipulation and coefficient comparison.
Standard integration formulae: It strengthens the understanding of logarithmic integration involving linear expressions in the denominator.
Stepwise mathematical presentation: Properly showing factorisation, decomposition, determination of constants and integration helps present a complete and logically justified solution.

Significance for Competitive Entrance Examinations

The concepts applied in this problem are useful for JEE Main, JEE Advanced, BITSAT and other engineering entrance examinations.

Recognition of algebraic identities: Quickly identifying difference of squares helps simplify rational expressions before integration.
Partial fraction techniques: This method is frequently useful in evaluating rational integrals involving distinct linear factors.
Standard integral recognition: Students should recognise expressions of the form 1/(x - a) and 1/(x + a) as logarithmic integrals.
Sign management: Correctly handling negative coefficients is important in objective questions where closely related answer choices may be given.
Equivalent answer forms: The ability to manipulate logarithmic expressions helps students identify mathematically equivalent options.
Advanced integration foundation: This technique prepares students for more complicated rational functions involving multiple linear factors, repeated factors and irreducible quadratic factors.

🚫 Common Mistakes

Common Mistakes

›

Incorrect factorisation: Remember that x² - 9 = (x - 3)(x + 3), not (x - 3)².
Incorrect partial fraction signs: The correct decomposition is 1/[6(x - 3)] - 1/[6(x + 3)].
Coefficient comparison errors: Since the numerator is 1, the coefficient of x must be zero and the constant term must be 1.
Incorrect determination of constants: From A + B = 0, we get A = -B, not A = B.
Missing numerical coefficient: The factor 1/6 must remain outside both logarithmic terms.
Incorrect logarithmic simplification: The difference of two logarithms becomes the logarithm of a quotient.
Omitting absolute values: Always write log|x - 3| and log|x + 3| in the indefinite integral.
Missing integration constant: Include C in the final expression.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 9 points

The difference of two squares identity is useful for factorising quadratic denominators.
A rational function with two distinct linear factors can be decomposed into two partial fractions.
The unknown constants in partial fractions can be calculated by comparing coefficients.
The fundamental logarithmic integration formula is:
\[ \boxed{ \int\frac{dx}{x-a}=\log|x-a|+C } \]
The difference of logarithms can be combined into the logarithm of a quotient.
Absolute value signs are necessary to ensure that the logarithmic expression is defined throughout the real domain of the integrand.
Differentiation provides a reliable method of verifying indefinite integrals.
Partial fraction decomposition is an essential technique for evaluating rational functions in Class 12 Mathematics and competitive entrance examinations.

📘 Concept & Theory

Theory and Concept

›

Integration of Rational Functions with Distinct Linear Factors

A rational function is an algebraic expression represented as the ratio of two polynomials. When the denominator is factorised into distinct linear factors and the degree of the numerator is less than the degree of the denominator, the function can be decomposed into simpler fractions using the method of partial fractions.

In this question, the denominator consists of three distinct linear factors:

\[ (x-1)(x-2)(x-3) \]

Therefore, the appropriate partial fraction decomposition is:

\[ \frac{P(x)}{(x-a)(x-b)(x-c)} = \frac{A}{x-a} +\frac{B}{x-b} +\frac{C}{x-c} \]

Here, A, B and C are constants determined by eliminating the denominators and comparing coefficients or substituting suitable values of x.

Important integration formula:

\[ \boxed{ \int\frac{1}{x-a}\,dx = \log|x-a|+C } \]

Consequently, once the rational function is decomposed into simpler fractions, each term can be integrated independently using the logarithmic integration formula.

Important observation:

The denominator has three distinct linear factors.
The numerator is a first-degree polynomial, whereas the denominator is a third-degree polynomial.
Since the rational function is proper, direct partial fraction decomposition is applicable.
Each distinct linear factor requires a separate constant numerator.
The coefficients A, B and C can be determined efficiently by substituting the roots of the denominator into the resulting identity.

🗺️ Solution Roadmap

Step-by-step Plan

›

Identify the three distinct linear factors in the denominator.
Assume a partial fraction decomposition with three unknown constants A, B and C.
Multiply both sides by the common denominator to eliminate fractions.
Expand the resulting expression and collect the coefficients of x², x and the constant terms.
Form three simultaneous equations and solve them to determine A, B and C.
Substitute the values of the constants into the original integral.
Separate the expression into three simpler integrals.
Apply the logarithmic integration formula and combine the results.
Verify the final answer by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 36 steps

Given —
\[I=\int\frac{3x-1}{(x-1)(x-2)(x-3)}\,dx\]
Apply partial fraction decomposition.
Since the denominator contains three distinct linear factors, assume:
\[\frac{3x-1}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}\]
Taking the LCM on the right-hand side:
\[ \begin{aligned} \frac{3x-1}{(x-1)(x-2)(x-3)} ={}& \frac{A(x-2)(x-3)}{(x-1)(x-2)(x-3)} \\[4pt] &+\frac{B(x-1)(x-3)}{(x-1)(x-2)(x-3)} \\[4pt] &+\frac{C(x-1)(x-2)}{(x-1)(x-2)(x-3)} \end{aligned} \]
Since the denominators on both sides are identical, equating the numerators gives:
\[ \begin{aligned} 3x-1={}&A(x-2)(x-3)\\ &+B(x-1)(x-3)\\ &+C(x-1)(x-2) \end{aligned} \]
Expand the products
Expanding each product separately:
\[(x-2)(x-3)=x^2-5x+6\]

\[(x-1)(x-3)=x^2-4x+3\]

\[(x-1)(x-2)=x^2-3x+2\]
Substituting these expansions into the numerator identity:
\[ \begin{aligned} 3x-1={}&A(x^2-5x+6)\\ &+B(x^2-4x+3)\\ &+C(x^2-3x+2) \end{aligned} \]
Multiplying each term by its respective constant:
\[ \begin{aligned} 3x-1={}&Ax^2-5Ax+6A\\ &+Bx^2-4Bx+3B\\ &+Cx^2-3Cx+2C \end{aligned} \]
Collect like terms
Grouping the coefficients of x², x and the constant terms:
\[ \begin{aligned} 3x-1={}&(A+B+C)x^2\\ &+(-5A-4B-3C)x\\ &+(6A+3B+2C) \end{aligned} \]
Compare coefficients
Comparing the coefficients of x² on both sides:
\[A+B+C=0 \tag{1}\]
Comparing the coefficients of x:
\[-5A-4B-3C=3 \tag{2}\]
Comparing the constant terms:
\[6A+3B+2C=-1 \tag{3}\]
We now have three simultaneous linear equations:
\[ \begin{align} A+B+C&=0 \tag{1}\\ -5A-4B-3C&=3 \tag{2}\\ 6A+3B+2C&=-1 \tag{3} \end{align} \]
Solve the simultaneous equations.
From equation (1):
\[A+B+C=0\]
Multiplying equation (1) by 3:
\[3A+3B+3C=0 \tag{4}\]
Subtracting equation (4) from equation (2):
\[ \begin{aligned} &(-5A-4B-3C)\\ &-(3A+3B+3C)=3 \end{aligned} \]
Simplifying:
\[-8A-7B-6C=3\]
Alternatively, using equation (1), substitute C = -A - B directly into equations (2) and (3).
From equation (1):
\[C=-A-B \tag{5}\]
Substituting equation (5) into equation (2):
\[-5A-4B-3(-A-B)=3\]
Removing the brackets:
\[-5A-4B+3A+3B=3\]
Combining like terms:
\[-2A-B=3\]
Multiplying both sides by -1:
\[2A+B=-3 \tag{6}\]
Now substituting equation (5) into equation (3):
\[6A+3B+2(-A-B)=-1\]
Removing the brackets:
\[6A+3B-2A-2B=-1\]
Combining like terms:
\[4A+B=-1 \tag{7}\]
Subtracting equation (6) from equation (7):
\[ \begin{aligned} (4A+B)-(2A+B)&=-1-(-3)\\ 4A+B-2A-B&=-1+3\\ 2A&=2 \end{aligned} \]
Dividing both sides by 2:
\[\boxed{A=1}\]
Substituting A = 1 into equation (6):
\[2(1)+B=-3\]

\[2+B=-3\]
Therefore:
\[\boxed{B=-5}\]
Finally, substituting A = 1 and B = -5 into equation (5):
\[C=-A-B\]

\[C=-1-(-5)\]
Hence:
\[\boxed{C=4}\]
Thus, the required constants are:
\[\boxed{A=1,\quad B=-5,\quad C=4}\]
Write the partial fraction decomposition
Substituting the values of A, B and C:
\[ \boxed{ \frac{3x-1}{(x-1)(x-2)(x-3)} = \frac{1}{x-1} -\frac{5}{x-2} +\frac{4}{x-3} } \]
Substitute the decomposition into the integral
We have:
\[I=\int\frac{3x-1}{(x-1)(x-2)(x-3)}\,dx\]
Using the partial fraction decomposition:
\[I=\int\left(\frac{1}{x-1}-\frac{5}{x-2}+\frac{4}{x-3}\right)dx\]
Applying the linearity property of integration:
\[ \begin{aligned} I={}&\int\frac{dx}{x-1}\\ &-5\int\frac{dx}{x-2}\\ &+4\int\frac{dx}{x-3} \end{aligned} \]
Apply the standard logarithmic integration formula.
\[\int\frac{dx}{x-a}=\log|x-a|+C\]
Integrating each term separately:
\[ \begin{aligned} I={}&\log|x-1|\\ &-5\log|x-2|\\ &+4\log|x-3|+C \end{aligned} \]
Therefore, the final result is:
\[ \boxed{ \begin{aligned} \displaystyle I={}&\log|x-1|\\ &-5\log|x-2|\\ &+4\log|x-3|+C \end{aligned} } \]

💡 Answer

Final Answer

›

🎯 Exam Significance

Exam Significance

›

This problem is important for CBSE Class 12 Mathematics board examination preparation because it demonstrates the integration of rational functions having three distinct linear factors in the denominator.

Application of partial fractions: It tests the ability to identify the correct decomposition when the denominator contains three distinct linear factors.
Polynomial expansion: Students practise expanding products of binomials and collecting coefficients of x², x and constant terms.
Simultaneous equations: Determining three unknown constants reinforces algebraic problem-solving skills.
Logarithmic integration: The problem strengthens the application of standard logarithmic integration formulae.
Mathematical presentation: Writing each algebraic transformation clearly helps produce a logically structured, examination-ready solution.
Answer verification: Differentiation confirms the correctness of the antiderivative and improves mathematical accuracy.

Significance for Competitive Entrance Examinations

The techniques used in this problem are relevant to JEE Main, JEE Advanced, BITSAT and other engineering entrance examinations.

Multiple linear factors: Students learn to handle rational functions with three or more distinct linear factors.
Efficient coefficient determination: The unknown constants can be calculated using algebraic comparison or the substitution method, helping reduce calculation time.
Substitution method: Substituting x = 1, 2 and 3 into the numerator identity provides a particularly efficient alternative for finding A, B and C.
Logarithmic identities: Combining several logarithmic terms into a single expression helps identify equivalent answer options.
Algebraic accuracy: Correct handling of positive and negative coefficients is essential in objective-type questions.
Advanced integration: This technique forms the foundation for evaluating more complicated rational integrals involving repeated linear factors and irreducible quadratic factors.

🚫 Common Mistakes

Common Mistakes

›

Incorrect partial fraction structure: For three distinct linear factors, use three separate constants A, B and C.
Incorrect expansion: Remember that (x - 2)(x - 3) = x² - 5x + 6.
Ignoring the x² coefficient: Since the numerator 3x - 1 has no x² term, the coefficient of x² in the expanded identity must be zero.
Sign errors: Pay particular attention to the negative coefficients of B and the signs in the constant terms.
Incorrect elimination: When substituting C = -A - B, carefully distribute the negative signs.
Missing integration coefficients: The constants -5 and 4 must remain as multipliers of their respective logarithmic terms.
Incorrect logarithmic simplification: A coefficient multiplying a logarithm becomes an exponent when the expression is combined into a single logarithm.
Omitting absolute values: Use absolute value signs in all logarithmic terms.
Forgetting the integration constant: Always include C in the final indefinite integral.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 8 points

Partial fraction decomposition is an effective technique for integrating proper rational functions.
A denominator containing three distinct linear factors requires three partial fractions with constant numerators.
The unknown constants can be determined by comparing coefficients or substituting the roots of the denominator.
When comparing coefficients, the coefficient of x² must be zero if the original numerator is a first-degree polynomial.
Each partial fraction with a linear denominator produces a logarithmic term upon integration.
The logarithmic expression can be simplified using the product, quotient and power properties of logarithms.
Differentiating the final expression is a reliable method of checking the answer.
The substitution method for finding partial fraction constants is particularly useful for saving time in competitive examinations.

📘 Concept & Theory

Theory and Concept

›

Integration of Rational Functions Having Distinct Linear Factors

A rational function is an algebraic expression represented as the ratio of two polynomials. When the degree of the numerator is less than the degree of the denominator, the rational function is called a proper rational function.

Partial fraction decomposition is an important technique for integrating proper rational functions. It involves expressing a complicated rational function as the sum of simpler rational expressions that can be integrated independently.

In the given question, the denominator consists of three distinct linear factors:

\[ (x-1)(x-2)(x-3) \]

Therefore, the appropriate partial fraction decomposition is:

\[ \frac{x}{(x-1)(x-2)(x-3)} = \frac{A}{x-1} +\frac{B}{x-2} +\frac{C}{x-3} \]

Here, A, B and C are constants that can be determined by comparing coefficients or substituting suitable values of x.

Important integration formula:

\[ \boxed{ \int\frac{dx}{x-a} = \log|x-a|+C } \]

Consequently, once the rational function is decomposed into partial fractions, each term can be integrated using the standard logarithmic integration formula.

Important observations:

The numerator is a polynomial of degree one.
The denominator is a polynomial of degree three.
Since the degree of the numerator is less than that of the denominator, the function is proper.
The denominator contains three distinct linear factors, so three partial fractions with constant numerators are required.
The coefficients of the partial fractions can be determined efficiently using coefficient comparison or the substitution method.

🗺️ Solution Roadmap

Step-by-step Plan

›

Identify the three distinct linear factors in the denominator.
Assume a partial fraction decomposition involving three unknown constants A, B and C.
Multiply both sides by the common denominator to eliminate fractions.
Expand the products and collect the coefficients of x², x and the constant terms.
Form three simultaneous equations and solve them to determine A, B and C.
Substitute the constants into the original integral.
Separate the resulting expression into three simpler integrals.
Apply the logarithmic integration formula to each term.
Verify the final answer by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 34 steps

Given —
\[I=\int\frac{x}{(x-1)(x-2)(x-3)}\,dx\]
Apply partial fraction decomposition
Since the denominator contains three distinct linear factors, assume:
\[\frac{x}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}\]
Taking the LCM on the right-hand side:
\[\begin{aligned}\frac{x}{(x-1)(x-2)(x-3)}={}&\frac{A(x-2)(x-3)}{(x-1)(x-2)(x-3)}\\[4pt] &+\frac{B(x-1)(x-3)}{(x-1)(x-2)(x-3)}\\[4pt] &+\frac{C(x-1)(x-2)}{(x-1)(x-2)(x-3)}\end{aligned}\]
Since the denominators on both sides are identical, equating the numerators gives:
\[ \begin{aligned} x={}&A(x-2)(x-3)\\ &+B(x-1)(x-3)\\ &+C(x-1)(x-2) \end{aligned} \]
Expand the products
Expanding each product separately:
\[(x-2)(x-3)=x^2-5x+6\]

\[(x-1)(x-3)=x^2-4x+3\]

\[(x-1)(x-2)=x^2-3x+2\]
Substituting these expansions into the numerator identity:
\[ \begin{aligned} x={}&A(x^2-5x+6)\\ &+B(x^2-4x+3)\\ &+C(x^2-3x+2) \end{aligned} \]
Multiplying each term by its respective constant:
\[ \begin{aligned} x={}&Ax^2-5Ax+6A\\ &+Bx^2-4Bx+3B\\ &+Cx^2-3Cx+2C \end{aligned} \]
Collect like terms
Grouping the coefficients of x², x and the constant terms:
\[ \begin{aligned} x={}&(A+B+C)x^2\\ &+(-5A-4B-3C)x\\ &+(6A+3B+2C) \end{aligned} \]
Compare coefficients
Comparing the coefficients of x² on both sides:
\[A+B+C=0 \tag{1}\]
Comparing the coefficients of x:
\[-5A-4B-3C=1 \tag{2}\]
Comparing the constant terms:
\[6A+3B+2C=0 \tag{3}\]
Thus, we obtain three simultaneous linear equations:
\[ \begin{align} A+B+C&=0 \tag{1}\\ -5A-4B-3C&=1 \tag{2}\\ 6A+3B+2C&=0 \tag{3} \end{align} \]
Solve the simultaneous equations.
From equation (1):
\[A+B+C=0\]
Multiplying equation (1) by 2:
\[2A+2B+2C=0 \tag{4}\]
Subtracting equation (4) from equation (3):
\[ \begin{aligned} &(6A+3B+2C)\\ &-(2A+2B+2C)=0 \end{aligned} \]
Simplifying:
\[4A+B=0 \tag{5}\]
Now, multiplying equation (1) by 3:
\[3A+3B+3C=0 \tag{6}\]
Adding equation (6) to equation (2):
\[ \begin{aligned} &(-5A-4B-3C)\\ &+(3A+3B+3C)=1 \end{aligned} \]
Simplifying:
\[-2A-B=1 \tag{7}\]
We now have two equations:
\[4A+B=0 \tag{5}\]

\[-2A-B=1 \tag{7}\]
Adding equations (5) and (7):
\[ \begin{aligned} (4A+B)+(-2A-B)&=0+1\\ 4A-2A+B-B&=1\\ 2A&=1 \end{aligned} \]
Dividing both sides by 2:
\[\boxed{A=\frac{1}{2}}\]
Substituting A = 1/2 into equation (5):
\[4A+B=0\]

\[4\left(\frac{1}{2}\right)+B=0\]

\[2+B=0\]
Therefore:
\[\boxed{B=-2}\]
Finally, substituting A = 1/2 and B = -2 into equation (1):
\[A+B+C=0\]

\[\frac{1}{2}-2+C=0\]
Taking the LCM:
\[\frac{1-4}{2}+C=0\]

\[-\frac{3}{2}+C=0\]
Hence:
\[\boxed{C=\frac{3}{2}}\]
Thus, the required constants are:
\[\boxed{A=\frac{1}{2},\quad B=-2,\quad C=\frac{3}{2}}\]
Write the partial fraction decomposition
Substituting the values of A, B and C:
\[ \boxed{ \frac{x}{(x-1)(x-2)(x-3)} = \frac{1}{2(x-1)} -\frac{2}{x-2} +\frac{3}{2(x-3)} } \]
Substitute the decomposition into the integral.
we have
\[I=\int\frac{x}{(x-1)(x-2)(x-3)}\,dx\]
Using the partial fraction decomposition:
\[I=\int\left[\frac{1}{2(x-1)}-\frac{2}{x-2}+\frac{3}{2(x-3)}\right]dx\]
Applying the linearity property of integration:
\[ \begin{aligned} I={}&\frac{1}{2}\int\frac{dx}{x-1}\\ &-2\int\frac{dx}{x-2}\\ &+\frac{3}{2}\int\frac{dx}{x-3} \end{aligned} \]
Apply the logarithmic integration formula.
\[\int\frac{dx}{x-a}=\log|x-a|+C\]
Integrating each term separately:
\[ \begin{aligned} I={}&\frac{1}{2}\log|x-1|\\ &-2\log|x-2|\\ &+\frac{3}{2}\log|x-3|+C \end{aligned} \]
Therefore, the final result is:
\[ \boxed{ \begin{aligned} \displaystyle I={}&\frac{1}{2}\log|x-1|\\ &-2\log|x-2|\\ &+\frac{3}{2}\log|x-3|+C \end{aligned} } \]

🎯 Exam Significance

Exam Significance

›

This problem is important for CBSE Class 12 Mathematics board examination preparation because it tests the application of partial fraction decomposition to a rational function with three distinct linear factors.

Understanding partial fractions: It develops the ability to decompose a rational function into simpler fractions before integration.
Algebraic manipulation: Students practise expanding polynomial products, collecting like terms and comparing coefficients.
Simultaneous linear equations: Determining three unknown constants reinforces the application of algebraic elimination techniques.
Logarithmic integration: It strengthens the understanding of standard logarithmic integration formulae.
Stepwise presentation: The solution demonstrates how to present the decomposition, calculations and integration in a clear mathematical sequence.
Answer verification: Differentiation provides an independent check of the final result.

Significance for Competitive Entrance Examinations

The concepts applied in this question are useful for JEE Main, JEE Advanced, BITSAT and other engineering entrance examinations.

Handling multiple linear factors: Students develop confidence in integrating rational functions containing three distinct linear factors.
Efficient calculation: The substitution method can determine partial fraction constants without lengthy polynomial expansion.
Coefficient comparison: Understanding the relationship between polynomial coefficients helps solve more advanced algebraic and integration problems.
Logarithmic simplification: Combining logarithmic terms helps identify equivalent answer choices in objective examinations.
Sign accuracy: Correct handling of positive and negative coefficients is essential for avoiding errors in multiple-choice questions.
Advanced integration foundation: This method prepares students for rational functions involving repeated factors and irreducible quadratic factors.

🚫 Common Mistakes

Common Mistakes

›

Incorrect partial fraction structure: Three distinct linear factors require three separate fractions with constants A, B and C.
Incorrect expansion: Remember that (x - 2)(x - 3) = x² - 5x + 6.
Incorrect coefficient comparison: The original numerator is x, so the coefficient of x² and the constant term must both be zero.
Sign errors during elimination: Pay particular attention when subtracting simultaneous equations.
Incorrect value of B: The coefficient of 1/(x - 2) is negative, namely -2.
Missing numerical coefficients: Retain 1/2 and 3/2 in the first and third logarithmic terms, respectively.
Incorrect logarithmic simplification: Fractional coefficients become fractional exponents when logarithmic terms are combined.
Omitting absolute values: Always include absolute value signs in logarithmic expressions.
Forgetting the integration constant: Include C in the final indefinite integral.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 8 points

Partial fraction decomposition is an effective technique for integrating proper rational functions.
Three distinct linear factors in the denominator require three partial fractions.
The unknown constants can be calculated using coefficient comparison or substitution of the roots of the denominator.
The original numerator determines the coefficients that must be compared during the decomposition.
Each fraction with a linear denominator produces a logarithmic term upon integration.
Logarithmic identities help combine multiple logarithmic terms into a single expression.
Differentiation is an effective method of verifying the correctness of an indefinite integral.
The substitution method is a valuable time-saving technique for objective-type competitive examinations.

📘 Concept & Theory

Theory and Concept

›

Integration of Rational Functions Using Factorisation and Partial Fractions

A rational function is an algebraic expression represented as the ratio of two polynomials. When the denominator can be factorised into distinct linear factors and the degree of the numerator is less than the degree of the denominator, partial fraction decomposition can be used to simplify the integration.

In this question, the denominator is a quadratic polynomial:

\[ x^2+3x+2 \]

We first factorise it by splitting the middle term. We need two numbers whose product is 2 and whose sum is 3. These numbers are 1 and 2.

Therefore:

\[ \begin{aligned} x^2+3x+2 &=x^2+2x+x+2\\ &=x(x+2)+1(x+2)\\ &=(x+1)(x+2) \end{aligned} \]

Since the denominator contains two distinct linear factors, the appropriate partial fraction decomposition is:

\[ \frac{2x}{(x+1)(x+2)} = \frac{A}{x+1}+\frac{B}{x+2} \]

Here, A and B are constants determined by comparing coefficients or substituting suitable values of x.

Important integration formula:

\[ \boxed{ \int\frac{dx}{x+a} = \log|x+a|+C } \]

👁️ Observation

Important observations

›

The numerator is a first-degree polynomial.
The denominator is a second-degree polynomial.
Since the numerator has a lower degree than the denominator, the rational function is proper.
The denominator can be factorised into two distinct linear factors.
After partial fraction decomposition, the integral can be evaluated using standard logarithmic integration formulae.

🗺️ Solution Roadmap

Step-by-step Plan

›

Factorise the quadratic denominator by splitting its middle term.
Express the rational function as the sum of two partial fractions with unknown constants A and B.
Multiply both sides by the common denominator to eliminate fractions.
Expand the numerator and compare the coefficients of x and the constant terms.
Solve the resulting simultaneous equations to determine A and B.
Substitute the constants into the original integral.
Separate the expression into two simpler integrals.
Apply the logarithmic integration formula and simplify the result.
Verify the final answer by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 28 steps

Given —
\[I=\int\frac{2x}{x^2+3x+2}\,dx\]
Factorise the denominator
The given quadratic expression is:
\[x^2+3x+2\]
Splitting the middle term 3x as 2x + x:
\[x^2+2x+x+2\]
Taking common factors from the first two and last two terms:
\[x(x+2)+1(x+2)\]
Taking (x + 2) common:
\[\boxed{x^2+3x+2=(x+1)(x+2)}\]
Substituting this factorisation into the integral:
\[I=\int\frac{2x}{(x+1)(x+2)}\,dx\]
Apply partial fraction decomposition
Since the denominator consists of two distinct linear factors, assume:
\[\frac{2x}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}\]
Taking the LCM on the right-hand side:
\[\frac{2x}{(x+1)(x+2)}=\frac{A(x+2)+B(x+1)}{(x+1)(x+2)}\]
Since the denominators on both sides are identical, equating the numerators gives:
\[2x=A(x+2)+B(x+1)\]
Expand and collect like terms
Expanding the right-hand side:
\[2x=Ax+2A+Bx+B\]
Rearranging and collecting the coefficients of x:
\[2x=(A+B)x+(2A+B)\]
Compare coefficients
Comparing the coefficients of x on both sides:
\[A+B=2 \tag{1}\]
Comparing the constant terms:
\[2A+B=0 \tag{2}\]
Thus, we obtain two simultaneous linear equations:
\[ \begin{align} A+B&=2 \tag{1}\\ 2A+B&=0 \tag{2} \end{align} \]
Determine the values of A and B
From equation (1):
\[A+B=2\]
Therefore:
\[B=2-A \tag{3}\]
Substituting equation (3) into equation (2):
\[2A+B=0\]

\[2A+(2-A)=0\]
Removing the brackets:
\[2A+2-A=0\]
Combining like terms:
\[A+2=0\]
Therefore:
\[boxed{A=-2}\]
Substituting A = -2 into equation (1):
\[-2+B=2\]
Adding 2 to both sides:
\[\boxed{B=4}\]
Thus, the required constants are:
\[\boxed{A=-2,\quad B=4}\]
Write the partial fraction decomposition.
Substituting the values of A and B:
\[\boxed{\frac{2x}{(x+1)(x+2)}=-\frac{2}{x+1}+\frac{4}{x+2}}\]
Substitute the decomposition into the integral
We have:
\[I=\int\frac{2x}{(x+1)(x+2)}\,dx\]
Using the partial fraction decomposition:
\[I=\int\left(-\frac{2}{x+1}+\frac{4}{x+2}\right)dx\]
Applying the linearity property of integration:
\[I=-2\int\frac{dx}{x+1}+4\int\frac{dx}{x+2}\]
Apply the standard logarithmic integration formula.
\[\int\frac{dx}{x+a}=\log|x+a|+C\]
Integrating each term separately:
\[I=-2\log|x+1|+4\log|x+2|+C\]
Therefore, the final result is:
\[\boxed{\displaystyle\int\frac{2x}{x^2+3x+2}\,dx=-2\log|x+1|+4\log|x+2|+C}\]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int\frac{2x}{x^2+3x+2}\,dx=-2\log|x+1|+4\log|x+2|+C}\]

🎯 Exam Significance

Exam Significance

›

This problem is important for CBSE Class 12 Mathematics board examination preparation because it combines quadratic factorisation, partial fraction decomposition and logarithmic integration.

Quadratic factorisation: It tests the ability to factorise a quadratic polynomial by splitting its middle term.
Partial fraction decomposition: Students learn to convert a rational function into simpler fractions with distinct linear denominators.
Algebraic accuracy: The determination of A and B reinforces coefficient comparison and the solution of simultaneous linear equations.
Standard integration formulae: It strengthens the application of logarithmic integration involving linear expressions in the denominator.
Stepwise mathematical presentation: The complete working demonstrates how to present factorisation, decomposition, determination of constants and integration systematically.
Verification: Differentiating the final answer provides an independent check of the result.

Significance for Competitive Entrance Examinations

The concepts used in this problem are relevant to JEE Main, JEE Advanced, BITSAT and other engineering entrance examinations.

Quick factorisation: Recognising suitable factors of quadratic expressions helps simplify rational functions efficiently.
Partial fraction techniques: The decomposition of rational functions into simpler fractions is useful in a wide range of integration problems.
Coefficient determination: Students develop the ability to determine unknown constants using algebraic identities and substitution.
Logarithmic integration: Recognising the standard form 1/(x + a) helps in identifying antiderivatives quickly.
Equivalent expressions: Logarithmic simplification helps students recognise mathematically equivalent answer options.
Advanced problem-solving: The same approach can be extended to rational functions with several distinct linear factors and more complicated denominators.

🚫 Common Mistakes

Common Mistakes

›

Incorrect factorisation: Remember that x² + 3x + 2 = (x + 1)(x + 2).
Incorrect partial fraction assumption: For two distinct linear factors, use A/(x + 1) + B/(x + 2).
Sign errors: Carefully expand A(x + 2) + B(x + 1) and retain the correct signs.
Incorrect coefficient comparison: The coefficient of x is 2, whereas the constant term is zero.
Incorrect values of constants: The correct values are A = -2 and B = 4.
Missing numerical coefficients: Retain -2 and 4 as multipliers of their respective logarithmic terms.
Incorrect logarithmic simplification: A numerical coefficient outside a logarithm becomes an exponent when the expression is combined into a single logarithm.
Omitting absolute values: Use log|x + 1| and log|x + 2| in the final expression.
Forgetting the integration constant: Always include C in an indefinite integral.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 10 points

Factorisation is often the first essential step in integrating rational functions.
A quadratic polynomial with two distinct linear factors can be decomposed into two partial fractions.
The unknown constants can be determined by comparing coefficients or substituting the roots of the denominator.
The standard logarithmic integration formula is:
\[ \boxed{ \int\frac{dx}{x+a}=\log|x+a|+C } \]
Each distinct linear factor in the denominator produces a separate logarithmic term.
Numerical coefficients multiplying logarithms become exponents when the terms are combined.
Absolute value signs are essential in logarithmic expressions involving variable quantities.
Differentiation provides a reliable method of verifying the correctness of an indefinite integral.
Mastering partial fraction decomposition is important for both board examinations and competitive entrance examinations.

📘 Concept & Theory

Theory and Concept

›

Integration of Rational Functions Using Polynomial Division and Partial Fractions

A rational function is an algebraic expression represented as the ratio of two polynomials. When the degree of the numerator is greater than or equal to the degree of the denominator, polynomial division may be required before applying partial fraction decomposition.

In this question, the numerator and denominator are:

\[ P(x)=1-x^2 \]

\[ Q(x)=x-2x^2 \]

Both polynomials have degree two. Therefore, the rational function is an improper rational function, and we first simplify it using polynomial division.

After polynomial division, the remaining proper rational function can be decomposed into partial fractions.

Important integration formulae:

\[ \boxed{\int dx=x+C} \]

\[ \boxed{\int\frac{dx}{x}=\log|x|+C} \]

\[ \boxed{ \int\frac{dx}{ax+b} = \frac{1}{a}\log|ax+b|+C } \]

For the last formula, the coefficient of x in the denominator must be included carefully. In particular:

\[ \boxed{ \int\frac{dx}{1-2x} = -\frac12\log|1-2x|+C } \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Identify the degrees of the numerator and denominator.
Since the rational function is improper, perform polynomial division to separate the polynomial part from the proper rational part.
Express the remaining proper rational function as the sum of two partial fractions with unknown constants A and B.
Eliminate the denominators and compare coefficients to determine A and B.
Substitute the constants into the original integral.
Separate the expression into three simpler integrals.
Apply the standard integration formulae, using substitution for the linear denominator 1 - 2x.
Simplify the logarithmic expression and write the final answer.
Verify the result by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 39 steps

Given —
\[I=\int\frac{1-x^2}{x-2x^2}\,dx\]
Identify the nature of the rational function
The numerator and denominator are both polynomials of degree two:
\[\deg(1-x^2)=2\]

\[\deg(x-2x^2)=2\]
Simplify the rational function using polynomial division
We have:
\[\frac{1-x^2}{x-2x^2}\]
We can express the numerator in the following form:
\[1-x^2=\frac12(x-2x^2)+\frac12(2-x)\]
Dividing both sides by x - 2x²:
\[\frac{1-x^2}{x-2x^2}=\frac12+\frac{2-x}{2(x-2x^2)}\]
Factorising the denominator:
\[x-2x^2=x(1-2x)\]
x-2x^2=x(1-2x)
Therefore,
\[\boxed{\frac{1-x^2}{x-2x^2}=\frac12+\frac{2-x}{2x(1-2x)}}\]
Apply partial fraction decomposition
We now decompose the remaining proper rational function:
\[\frac{2-x}{2x(1-2x)}\]
Since the denominator contains two distinct linear factors, assume:
\[\frac{2-x}{2x(1-2x)}=\frac{A}{2x}+\frac{B}{1-2x}\]
Taking the LCM on the right-hand side:
\[\frac{2-x}{2x(1-2x)}=\frac{A(1-2x)+2Bx}{2x(1-2x)}\]
Since the denominators are identical, equating the numerators gives:
\[2-x=A(1-2x)+2Bx\]
Expand and collect like terms
Expanding the right-hand side:
\[2-x=A-2Ax+2Bx\]
Collecting the coefficients of x:
\[2-x=A+(-2A+2B)x\]
Compare coefficients
Comparing the constant terms:
\[A=2 \tag{1}\]
Comparing the coefficients of x:
\[-2A+2B=-1 \tag{2}\]
Determine the values of A and B
From equation (1):
\[\boxed{A=2}\]
Substituting A = 2 into equation (2):
\[-2(2)+2B=-1\]
Simplifying:
\[-4+2B=-1\]
Adding 4 to both sides:
\[2B=3\]
Dividing both sides by 2:
\[\boxed{B=\frac32}\]
Hence, the required constants are:
\[\boxed{A=2,\quad B=\frac32}\]
Write the partial fraction decomposition.
Substituting the values of A and B:
\[\frac{2-x}{2x(1-2x)}=\frac{2}{2x}+\frac{3}{2(1-2x)}\]
Simplifying the first term:
\[\boxed{\frac{2-x}{2x(1-2x)}=\frac1x+\frac{3}{2(1-2x)}}\]
Therefore, the complete decomposition of the original rational function is:
\[\boxed{\frac{1-x^2}{x-2x^2}=\frac12+\frac1x+\frac{3}{2(1-2x)}}\]
Substitute the decomposition into the integral.
We have:
\[I=\int\frac{1-x^2}{x-2x^2}\,dx\]
Using the simplified expression:
\[I=\int\left(\frac12+\frac1x+\frac{3}{2(1-2x)}\right)dx\]
Applying the linearity property of integration:
\[I=\frac12\int dx+\int\frac{dx}{x}+\frac32\int\frac{dx}{1-2x}\]
Evaluate the individual integrals
First integral:
\[\frac12\int dx=\frac{x}{2}\]
Second integral:
\[\int\frac{dx}{x}=\log|x|\]
Third integral:
We need to evaluate:
\[\int\frac{dx}{1-2x}\]
Using substitution, let:
\[t=1-2x\]
Differentiating both sides:
\[\frac{dt}{dx}=-2\]
Therefore:
\[dt=-2\,dx\]
Rearranging:
\[dx=-\frac12\,dt\]
Substituting into the integral:
\[ \begin{aligned} \int\frac{dx}{1-2x} &=-\frac12\int\frac{dt}{t}\\ &=-\frac12\log|t|+C\\ &=-\frac12\log|1-2x|+C \end{aligned} \]
Hence:
\[\boxed{\int\frac{dx}{1-2x}=-\frac12\log|1-2x|+C}\]
Combine the results
Substituting the three evaluated integrals:
\[I=\frac{x}{2}+\log|x|+\frac32\left(-\frac12\log|1-2x|\right)+C\]
Multiplying the coefficients:
\[I=\frac{x}{2}+\log|x|-\frac34\log|1-2x|+C\]
Therefore, the final answer is:
\[\boxed{\displaystyle\int\frac{1-x^2}{x-2x^2}\,dx=\frac{x}{2}+\log|x|-\frac34\log|1-2x|+C}\]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int\frac{1-x^2}{x-2x^2}\,dx=\frac{x}{2}+\log|x|-\frac34\log|1-2x|+C}\]

🎯 Exam Significance

Exam Significance

›

This problem is important for CBSE Class 12 Mathematics board examination preparation because it combines polynomial division, partial fraction decomposition and logarithmic integration.

Identification of improper rational functions: Students learn to recognise when polynomial division is necessary before applying partial fractions.
Polynomial division: It tests the ability to separate a rational function into a polynomial part and a proper rational part.
Partial fraction decomposition: Students practise decomposing expressions containing distinct linear factors.
Algebraic accuracy: The determination of unknown constants reinforces coefficient comparison and simultaneous equations.
Integration by substitution: The term 1/(1 - 2x) demonstrates the importance of accounting for the derivative of the inner linear expression.
Stepwise presentation: The solution illustrates how to organise polynomial division, decomposition and integration into a complete mathematical argument.

Significance for Competitive Entrance Examinations

The concepts applied in this problem are useful for JEE Main, JEE Advanced, BITSAT and other engineering entrance examinations.

Improper rational functions: Recognising when polynomial division is required is an important preliminary step in integration problems.
Algebraic simplification: Efficient decomposition reduces complicated rational expressions into standard forms.
Partial fraction techniques: Students develop the ability to integrate rational functions with distinct linear factors.
Chain rule awareness: The negative coefficient in the derivative of 1 - 2x is crucial for obtaining the correct logarithmic integral.
Sign management: Negative signs in substitution and logarithmic integration are common sources of errors in objective examinations.
Verification skills: Differentiating an antiderivative is useful for checking answers and identifying incorrect options.

🚫 Common Mistakes

Common Mistakes

›

Skipping polynomial division: Since the numerator and denominator have equal degrees, first separate the polynomial part before applying partial fractions.
Incorrect denominator factorisation: Remember that x - 2x² = x(1 - 2x).
Incorrect partial fraction decomposition: The proper decomposition is A/(2x) + B/(1 - 2x).
Sign errors while comparing coefficients: The coefficient equation is -2A + 2B = -1.
Incorrect coefficient values: The correct values are A = 2 and B = 3/2.
Ignoring the chain rule: The derivative of 1 - 2x is -2, so its reciprocal integration introduces a negative factor of 1/2.
Incorrect logarithmic coefficient: The final coefficient of log|1 - 2x| is -3/4, not +3/2.
Omitting absolute values: Include absolute value signs in logarithmic expressions.
Forgetting the integration constant: Always include C in the final indefinite integral.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 10 points

An improper rational function must first be simplified using polynomial division before applying partial fraction decomposition.
The denominator x - 2x² can be factorised as x(1 - 2x).
The proper rational part can be expressed using two distinct linear partial fractions.
The unknown constants can be determined by coefficient comparison or suitable substitution.
When integrating a linear expression in the denominator, its derivative must be accounted for.
The essential integration formula is:
\[ \boxed{ \int\frac{dx}{ax+b} = \frac1a\log|ax+b|+C } \]
The negative coefficient in the derivative of 1 - 2x produces a negative logarithmic term.
Differentiation provides an effective method of verifying the final result.
Polynomial division and partial fraction decomposition are important techniques for solving rational integration problems.

📘 Concept & Theory

Theory and Concept

›

Integration of Rational Functions Containing Linear and Irreducible Quadratic Factors

A rational function is an algebraic expression represented as the ratio of two polynomials. When the denominator contains a combination of distinct linear factors and irreducible quadratic factors, partial fraction decomposition is used to simplify the rational function before integration.

In this question, the denominator contains two different types of factors:

1. A distinct linear factor:

\[x-1\]

2. An irreducible quadratic factor:

\[x^2+1\]

The quadratic expression x² + 1 cannot be factorised into real linear factors because its discriminant is negative.

According to the rules of partial fraction decomposition:

A distinct linear factor requires a constant numerator.
An irreducible quadratic factor requires a linear numerator of the form Bx + C.

Therefore, the appropriate partial fraction decomposition is:

\[\boxed{\frac{x}{(x-1)(x^2+1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+1}}\]

Here, A, B and C are constants determined by comparing coefficients.

Important integration formulae:

Logarithmic integration:

\[\boxed{\int\frac{dx}{x-a}=\log|x-a|+C}\]

Integration involving the derivative of a quadratic expression:

\[\boxed{\int\frac{x\,dx}{x^2+a^2}=\frac12\log(x^2+a^2)+C}\]

Inverse trigonometric integration:

\[\boxed{\int\frac{dx}{x^2+a^2}=\frac1a\tan^{-1}\left(\frac{x}{a}\right)+C}\]

For a = 1:

\[\boxed{\int\frac{dx}{x^2+1}=\tan^{-1}x+C}\]

Important observation:

When the numerator of a rational expression containing an irreducible quadratic factor is linear, it can be separated into two parts: one proportional to the derivative of the quadratic denominator and another constant term. These parts lead to logarithmic and inverse trigonometric integrals, respectively.

🗺️ Solution Roadmap

Step-by-step Plan

›

Identify the distinct linear factor and the irreducible quadratic factor in the denominator.
Assume the appropriate partial fraction decomposition with constants A, B and C.
Multiply both sides by the common denominator to eliminate fractions.
Expand the numerator and collect the coefficients of x², x and the constant terms.
Compare coefficients and solve the resulting simultaneous equations.
Substitute the values of A, B and C into the original integral.
Separate the resulting expression into three simpler integrals.
Evaluate the first integral using the logarithmic formula, the second using substitution and the third using the inverse tangent formula.
Combine the results and verify the final answer by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 44 steps

Given —
\[I=\int\frac{x}{(x-1)(x^2+1)}\,dx\]
Apply partial fraction decomposition.
The denominator contains one distinct linear factor (x - 1) and one irreducible quadratic factor (x² + 1).
Therefore, assume:
\[\frac{x}{(x-1)(x^2+1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+1}\]
Taking the LCM on the right-hand side:
\[\frac{x}{(x-1)(x^2+1)}=\frac{A(x^2+1)+(x-1)(Bx+C)}{(x-1)(x^2+1)}\]
Since the denominators on both sides are identical, equating the numerators gives:
\[x=A(x^2+1)+(x-1)(Bx+C)\]
Expand the products
First, expand the second product:
\[(x-1)(Bx+C)=Bx^2+Cx-Bx-C\]
Substituting into the numerator identity:
\[x=Ax^2+A+Bx^2+Cx-Bx-C\]
Collect like terms
Grouping the coefficients of x², x and the constant terms:
\[x=(A+B)x^2+(C-B)x+(A-C)\]
Compare coefficients.
Comparing the coefficients of x² on both sides:
\[A+B=0 \tag{1}\]
Comparing the coefficients of x:
\[C-B=1 \tag{2}\]
Comparing the constant terms:
\[A-C=0 \tag{3}\]
Thus, we obtain three simultaneous equations:
\[ \begin{align} A+B&=0 \tag{1}\\ C-B&=1 \tag{2}\\ A-C&=0 \tag{3} \end{align} \]
Determine the values of A, B and C.
From equation (1):
\[A+B=0\]
Therefore:
\[A=-B \tag{4}\]
From equation (3):
\[A-C=0\]
Therefore:
\[C=A \tag{5}\]
Substituting equations (4) and (5) into equation (2):
\[C-B=1\]

\[ A-B=1\]
Since A = -B:
\[-B-B=1\]
Combining like terms:
\[-2B=1\]
Dividing both sides by -2:
\[\boxed{B=-\frac12}\]
Using equation (4):
\[A=-B\]

\[A=-\left(-\frac12\right)\]
Therefore:
\[\boxed{A=\frac12}\]
Using equation (5):
\[C=A\]
Hence:
\[\boxed{C=\frac12}\]
Thus, the required constants are:
\[\boxed{A=\frac12,\quad B=-\frac12,\quad C=\frac12}\]
Write the partial fraction decomposition.
Substituting the values of A, B and C:
\[\frac{x}{(x-1)(x^2+1)}=\frac{1}{2(x-1)}+\frac{-\frac{x}{2}+\frac12}{x^2+1}\]
Separating the terms in the numerator:
\[\boxed{\frac{x}{(x-1)(x^2+1)}=\frac{1}{2(x-1)}-\frac{x}{2(x^2+1)}+\frac{1}{2(x^2+1)}}\]
Substitute the decomposition into the integral
We have:
\[I=\int\frac{x}{(x-1)(x^2+1)}\,dx\]
Using the partial fraction decomposition:
\[I=\int\left[\frac{1}{2(x-1)}-\frac{x}{2(x^2+1)}+\frac{1}{2(x^2+1)}\right]dx\]
Applying the linearity property of integration:
\[ \begin{aligned} I={}& \frac12\int\frac{dx}{x-1}\\ &-\frac12\int\frac{x\,dx}{x^2+1}\\ &+\frac12\int\frac{dx}{x^2+1} \end{aligned} \]
Evaluate the first integral
Consider:
\[\frac12\int\frac{dx}{x-1}\]
Using the standard logarithmic integration formula:
\[\int\frac{dx}{x-a}=\log|x-a|+C\]
We obtain:
\[\boxed{\frac12\int\frac{dx}{x-1}=\frac12\log|x-1|}\]
Evaluate the second integral using substitution
Consider:
\[-\frac12\int\frac{x\,dx}{x^2+1}\]
Let:
\[t=x^2+1\]
Differentiating both sides with respect to x:
\[\frac{dt}{dx}=2x\]
Therefore:
\[dt=2x\,dx\]
Rearranging:
\[x\,dx=\frac{dt}{2}\]
Substituting into the integral:
\[ \begin{aligned} -\frac12\int\frac{x\,dx}{x^2+1} &=-\frac12\int\frac{dt/2}{t}\\ &=-\frac14\int\frac{dt}{t}\\ &=-\frac14\log|t|+C \end{aligned} \]
Substituting t = x² + 1:
\[-\frac14\log|x^2+1|+C\]
Therefore:
\[\boxed{-\frac12\int\frac{x\,dx}{x^2+1}=-\frac14\log(x^2+1)}\]
Evaluate the third integral using the inverse tangent formula
Consider:
\[\frac12\int\frac{dx}{x^2+1}\]
\[\int\frac{dx}{x^2+a^2}=\frac1a\tan^{-1}\left(\frac{x}{a}\right)+C\]
Comparing the denominator x² + 1 with x² + a², we get:
\[a^2=1\]
Therefore:
\[a=1\]
Substituting a = 1 into the standard formula:
\[\int\frac{dx}{x^2+1}=\tan^{-1}x+C\]
Hence:
\[\boxed{\frac12\int\frac{dx}{x^2+1}=\frac12\tan^{-1}x}\]
Combine all the integrals.
We have obtained:
\[ \begin{aligned} I={}& \frac12\log|x-1|\\ &-\frac14\log(x^2+1)\\ &+\frac12\tan^{-1}x+C \end{aligned} \]

💡 Answer

Final Answer

›

\[ \boxed{ \begin{aligned} \displaystyle \int\frac{x}{(x-1)(x^2+1)}\,dx ={}& \frac12\log|x-1|\\ &-\frac14\log(x^2+1)\\ &+\frac12\tan^{-1}x+C \end{aligned} } \]

🎯 Exam Significance

Exam Significance

›

This problem is important for CBSE Class 12 Mathematics board examination preparation because it demonstrates the application of partial fractions involving both distinct linear factors and irreducible quadratic factors.

Partial fraction decomposition: It tests the ability to select the correct form of decomposition for a denominator containing linear and irreducible quadratic factors.
Coefficient comparison: Students practise expanding polynomial expressions and determining three unknown constants.
Substitution method: The integral involving x/(x² + 1) reinforces the use of substitution when the numerator is proportional to the derivative of the denominator.
Inverse trigonometric integration: It tests the recognition and application of the standard integral involving 1/(x² + a²).
Multiple integration techniques: The problem combines logarithmic integration, substitution and inverse trigonometric integration in one solution.
Stepwise presentation: Showing the decomposition and evaluating each resulting integral separately makes the solution logically complete.

Significance for Competitive Entrance Examinations

The concepts used in this problem are relevant to JEE Main, JEE Advanced, BITSAT and other engineering entrance examinations.

Recognition of irreducible quadratic factors: Students learn to use a linear numerator when decomposing fractions involving irreducible quadratic denominators.
Mixed integration techniques: Competitive examinations frequently test the ability to identify different standard integral forms within a single expression.
Derivative-based substitution: Recognising the relationship between x and the derivative of x² + 1 helps simplify logarithmic integrals.
Inverse trigonometric functions: Familiarity with the standard arctangent integral is essential for solving a variety of integration problems.
Algebraic accuracy: Correctly determining the coefficients A, B and C prevents errors in the subsequent integration.
Answer verification: Differentiation helps identify incorrect signs and coefficients in multiple-choice solutions.

🚫 Common Mistakes

Common Mistakes

›

Incorrect partial fraction structure: For an irreducible quadratic factor, use a linear numerator Bx + C rather than a constant numerator alone.
Incorrect expansion: Remember that (x - 1)(Bx + C) = Bx² + (C - B)x - C.
Coefficient comparison errors: The correct equations are A + B = 0, C - B = 1 and A - C = 0.
Incorrect values of constants: The correct values are A = 1/2, B = -1/2 and C = 1/2.
Sign error in substitution: When t = x² + 1, we obtain x dx = dt/2.
Incorrect logarithmic coefficient: The integral of x/(x² + 1) is one-half log(x² + 1), so the resulting coefficient is -1/4.
Incorrect inverse tangent expression: Since a = 1, the correct expression is tan⁻¹x, not tan⁻¹(x/a) with an unspecified a.
Missing inverse tangent coefficient: Retain the factor 1/2 multiplying tan⁻¹x.
Forgetting the integration constant: Include C in the final indefinite integral.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 9 points

A denominator containing a distinct linear factor and an irreducible quadratic factor requires different numerator forms in partial fraction decomposition.
A linear factor requires a constant numerator, whereas an irreducible quadratic factor requires a linear numerator.
The constants in partial fractions can be determined through coefficient comparison or suitable substitution.
The integral of x/(x² + a²) is evaluated using substitution and produces a logarithmic expression.
The integral of 1/(x² + a²) produces an inverse tangent function.
When a = 1, the standard inverse tangent integral becomes tan⁻¹x.
Careful handling of coefficients and signs is essential when combining different integration techniques.
Differentiation provides an effective method of verifying the final antiderivative.
Mastering partial fractions involving irreducible quadratic factors is important for advanced integration problems in board and competitive examinations.

📘 Concept & Theory

Theory and Concept

›

Integration of Rational Functions Containing Repeated Linear Factors

A rational function is an algebraic expression represented as the ratio of two polynomials. When the denominator contains repeated linear factors, partial fraction decomposition is used to express the rational function as a sum of simpler fractions.

In this question, the denominator contains one repeated linear factor and one distinct linear factor:

\[ (x-1)^2(x+2) \]

The factor (x - 1) is repeated twice, whereas (x + 2) is a distinct linear factor.

Rule for repeated linear factors:

If the denominator contains a repeated linear factor (x - a)², its partial fraction decomposition must include a separate fraction for each power of that factor.

For example:

\[\frac{P(x)}{(x-a)^2(x-b)}=\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{(x-a)^2}\]

Thus, the appropriate partial fraction decomposition for the given problem is:

\[\boxed{\frac{x}{(x-1)^2(x+2)}=\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{(x-1)^2}}\]

Here, A, B and C are constants determined by eliminating the denominators and comparing coefficients.

Important integration formulae:

Logarithmic integration:

\[\boxed{\int\frac{dx}{x-a}=\log|x-a|+C}\]

Integration of a negative power:

\[\boxed{\int x^n\,dx=\frac{x^{n+1}}{n+1}+C,\quad n\neq-1}\]

In particular:

\[\boxed{\int\frac{dx}{(x-a)^2}=-\frac{1}{x-a}+C}\]

Important observation:

Unlike the integration of a simple linear factor, integrating a repeated linear factor with denominator (x - a)² produces a rational expression rather than a logarithmic expression.

🗺️ Solution Roadmap

Step-by-step Plan

›

Identify the repeated linear factor and the distinct linear factor in the denominator.
Assume the appropriate partial fraction decomposition involving three unknown constants A, B and C.
Multiply both sides by the common denominator to eliminate fractions.
Expand the products and collect the coefficients of x², x and the constant terms.
Form three simultaneous equations and solve them to determine A, B and C.
Substitute the constants into the original integral.
Separate the expression into three simpler integrals.
Evaluate the first two integrals using logarithmic integration and the third using the power rule of integration.
Combine the results and verify the final answer by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 49 steps

Given —
\[I=\int\frac{x}{(x-1)^2(x+2)}\,dx\]
Apply partial fraction decomposition
Since the denominator contains the repeated linear factor (x - 1)² and the distinct linear factor (x + 2), assume:
\[\frac{x}{(x-1)^2(x+2)}=\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{(x-1)^2}\]
Taking the LCM on the right-hand side:
\[\begin{aligned}\frac{x}{(x-1)^2(x+2)}=\frac{A(x-1)(x+2)+B(x-1)^2+C(x+2)}{(x-1)^2(x+2)}\end{aligned}\]
Since the denominators on both sides are identical, equating the numerators gives:
\[x=A(x-1)(x+2)+B(x-1)^2+C(x+2)\]
Expand the products
First, expand each product separately:
\[(x-1)(x+2)=x^2+x-2\]

\[(x-1)^2=x^2-2x+1\]
Substituting these expansions into the numerator identity:
\[x=A(x^2+x-2)+B(x^2-2x+1)+C(x+2)\]
Multiplying each term by its respective constant:
\[ \begin{aligned} x={}&Ax^2+Ax-2A\\ &+Bx^2-2Bx+B\\ &+Cx+2C \end{aligned} \]
Collect like terms
Grouping the coefficients of x², x and the constant terms:
\[x=(A+B)x^2+(A-2B+C)x+(-2A+B+2C)\]
Compare Coefficients
Comparing the coefficients of x² on both sides:
\[A+B=0 \tag{1}\]
Comparing the coefficients of x:
\[A-2B+C=1 \tag{2}\]
Comparing the constant terms:
\[-2A+B+2C=0 \tag{3}\]
Thus, we obtain three simultaneous equations:
\[ \begin{align} A+B&=0 \tag{1}\\ A-2B+C&=1 \tag{2}\\ -2A+B+2C&=0 \tag{3} \end{align} \]
Determine the values of A, B and C.
From equation (1):
\[A+B=0\]
Therefore:
\[B=-A \tag{4}\]
Substituting equation (4) into equation (3):
\[-2A+B+2C=0\]

\[-2A-A+2C=0\]
Combining like terms:
\[-3A+2C=0\]
Therefore:
\[2C=3A\]
Or:
\[C=\frac{3A}{2} \tag{5}\]
Now substitute equation (4) into equation (2):
\[A-2B+C=1\]

\[A-2(-A)+C=1\]
Removing the brackets:
\[A+2A+C=1\]
Combining like terms:
\[3A+C=1 \tag{6}\]
Substituting equation (5) into equation (6):
\[3A+\frac{3A}{2}=1\]
Taking the LCM:
\[\frac{6A+3A}{2}=1\]

\[\frac{9A}{2}=1\]
Multiplying both sides by 2:
\[9A=2\]
Therefore:
\[\boxed{A=\frac29}\]
Using equation (4):
\[B=-A\]
Hence:
\[\boxed{B=-\frac29}\]
Using equation (5):
\[C=\frac{3A}{2}\]
Substituting A = 2/9:
\[C=\frac32\times\frac29\]
Therefore:
\[\boxed{C=\frac13}\]
Thus, the required constants are:
\[\boxed{A=\frac29,\quad B=-\frac29,\quad C=\frac13}\]
Write the partial fraction decomposition.
Substituting the values of A, B and C:
\[\boxed{\frac{x}{(x-1)^2(x+2)}=\frac{2}{9(x-1)}-\frac{2}{9(x+2)}+\frac{1}{3(x-1)^2}}\]
Substitute the decomposition into the integral
We have:
\[I=\int\frac{x}{(x-1)^2(x+2)}\,dx\]
Using the partial fraction decomposition:
\[I=\int\left[\frac{2}{9(x-1)}-\frac{2}{9(x+2)}+\frac{1}{3(x-1)^2}\right]dx\]
Applying the linearity property of integration:
\[ \begin{aligned} I={}& \frac29\int\frac{dx}{x-1}\\ &-\frac29\int\frac{dx}{x+2}\\ &+\frac13\int\frac{dx}{(x-1)^2} \end{aligned} \]
Evaluate the first integral
Consider:
\[\frac29\int\frac{dx}{x-1}\]
Using the standard logarithmic integration formula:
\[\int\frac{dx}{x-a}=\log|x-a|+C\]
We obtain:
\[\boxed{\frac29\int\frac{dx}{x-1}=\frac29\log|x-1|}\]
Evaluate the second integral
Consider:
\[-\frac29\int\frac{dx}{x+2}\]
Applying the logarithmic integration formula:
\[\boxed{-\frac29\int\frac{dx}{x+2}=-\frac29\log|x+2|}\]
Evaluate the third integral using substitution
Consider:
\[\frac13\int\frac{dx}{(x-1)^2}\]
Let:
\[t=x-1\]
Differentiating both sides:
\[dt=dx\]
Substituting into the integral:
\[\frac13\int\frac{dt}{t^2}\]
Expressing the negative power:
\[\frac13\int t^{-2}\,dt\]
Using the power rule of integration:
\[\int t^n\,dt=\frac{t^{n+1}}{n+1}+C,\quad n\neq-1\]
Putting n = -2:
\[\begin{aligned}\frac13\int t^{-2}\,dt&=\frac13\left(\frac{t^{-2+1}}{-2+1}\right)\\ &=\frac13\left(\frac{t^{-1}}{-1}\right)\\ &=-\frac13t^{-1}\\ &=-\frac{1}{3t}\end{aligned}\]
Substituting t = x - 1:
\[\boxed{\frac13\int\frac{dx}{(x-1)^2}=-\frac{1}{3(x-1)}}\]
Combine all the integrals.
\[\begin{aligned}I={}&\frac29\log|x-1|\\ &-\frac29\log|x+2|\\ &-\frac{1}{3(x-1)}+C\end{aligned}\]
Using the logarithmic identity:
\[\log a-\log b=\log\left(\frac ab\right)\]
We obtain:
\[I=\frac29\log\left|\frac{x-1}{x+2}\right|-\frac{1}{3(x-1)}+C\]
Therefore,
\[\boxed{\displaystyle\int\frac{x}{(x-1)^2(x+2)}\,dx=\frac29\log\left|\frac{x-1}{x+2}\right|-\frac{1}{3(x-1)}+C}\]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int\frac{x}{(x-1)^2(x+2)}\,dx=\frac29\log\left|\frac{x-1}{x+2}\right|-\frac{1}{3(x-1)}+C}\]

🎯 Exam Significance

Exam Significance

›

This problem is important for CBSE Class 12 Mathematics board examination preparation because it demonstrates the integration of rational functions containing repeated linear factors.

Repeated linear factors: It tests the ability to construct the correct partial fraction form when a denominator contains a repeated factor.
Algebraic manipulation: Students practise polynomial expansion, coefficient comparison and solving simultaneous equations.
Power rule of integration: It reinforces the integration of negative powers, particularly the case involving (x - 1)⁻².
Logarithmic integration: Students learn to integrate the simple linear factors separately.
Sign accuracy: The problem highlights the importance of handling negative powers correctly during integration.
Stepwise presentation: It demonstrates how to organise partial fraction decomposition and integration into a complete examination-ready solution.

Significance for Competitive Entrance Examinations

The concepts applied in this question are useful for JEE Main, JEE Advanced, BITSAT and other engineering entrance examinations.

Repeated-factor decomposition: Students learn to handle rational functions with repeated linear factors in the denominator.
Efficient coefficient determination: Understanding the relationship between coefficients helps reduce unnecessary calculations.
Negative power integration: Recognising the power rule for negative exponents is essential for simplifying rational integrals.
Algebraic accuracy: Correct handling of signs and fractions is important in objective-type questions.
Equivalent expressions: Logarithmic simplification helps identify equivalent answer options.
Advanced integration foundation: This technique prepares students for more complicated rational functions involving repeated linear and quadratic factors.

🚫 Common Mistakes

Common Mistakes

›

Incorrect partial fraction structure: A repeated factor (x - 1)² requires both A/(x - 1) and C/(x - 1)².
Omitting a repeated-factor term: Do not represent the repeated factor using only one partial fraction.
Incorrect expansion: Remember that (x - 1)(x + 2) = x² + x - 2.
Incorrect coefficient comparison: The correct equations are A + B = 0, A - 2B + C = 1 and -2A + B + 2C = 0.
Sign errors during elimination: Carefully handle B = -A when substituting into the remaining equations.
Incorrect power rule application: The integral of t⁻² is -t⁻¹, not t⁻¹.
Missing negative sign: The integral of 1/(x - 1)² is -1/(x - 1), not +1/(x - 1).
Incorrect logarithmic simplification: The difference of logarithms becomes the logarithm of a quotient.
Forgetting the integration constant: Always include C in the final indefinite integral.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 11 points

A repeated linear factor in the denominator requires a separate partial fraction for every power of that factor.
The decomposition of a rational function with (x - 1)² and (x + 2) is different from that of a function with three distinct linear factors.
The unknown constants can be determined through coefficient comparison or suitable substitution.
Simple linear denominators produce logarithmic terms upon integration.
A repeated linear denominator with power two produces a rational term after integration.
The important formula for a repeated linear factor is:
\[\boxed{\int\frac{dx}{(x-a)^2}=-\frac{1}{x-a}+C}\]
Negative powers must be integrated carefully using the power rule.
Sign accuracy is essential when evaluating partial fractions and combining the resulting integrals.
Differentiation is an effective method of verifying the final antiderivative.
Mastering repeated linear factors provides a foundation for more advanced rational integration problems.

📘 Concept & Theory

Theory and Concept

›

Integration of Rational Functions Using Partial Fractions

A rational function is a quotient of two polynomials. When its denominator can be factorised into linear or irreducible quadratic factors, partial fraction decomposition can be used to express the function as a sum of simpler rational expressions.

The first step is to factorise the denominator completely. Consider the denominator in the given integral:

\[ x^3-x^2-x+1 \]

By grouping the terms:

\[\begin{aligned}x^3-x^2-x+1&=x^2(x-1)-1(x-1)\\ &=(x-1)(x^2-1)\\ &=(x-1)(x-1)(x+1)\end{aligned}\]

Therefore:

\[\boxed{x^3-x^2-x+1=(x-1)^2(x+1)}\]

The denominator contains two types of linear factors:

A repeated linear factor: (x - 1)².
A distinct linear factor: (x + 1).

Rule for partial fractions involving repeated linear factors:

If a denominator contains a repeated linear factor \((x-a)^2\), its partial fraction decomposition must include a term for each power of that factor:

\[\frac{P(x)}{(x-a)^2(x-b)}=\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{(x-a)^2}\]

Consequently, the appropriate decomposition for this question is:

\[\boxed{\frac{3x+5}{(x-1)^2(x+1)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{(x-1)^2}}\]

Here, A, B and C are constants to be determined.

Important integration formulae:

For a simple linear factor:

\[\int\frac{dx}{x-a}=\log|x-a|+C\]

For a repeated linear factor:

\[\int\frac{dx}{(x-a)^2}=-\frac{1}{x-a}+C\]

The first formula produces a logarithmic term, whereas the second produces a rational expression.

🗺️ Solution Roadmap

Step-by-step Plan

›

Factorise the cubic polynomial in the denominator completely.
Identify the repeated and distinct linear factors.
Assume the correct partial fraction decomposition.
Multiply both sides by the common denominator to eliminate fractions.
Expand the expressions and collect the coefficients of x², x and the constant term.
Form simultaneous equations and determine A, B and C.
Substitute the constants into the integral.
Evaluate the resulting logarithmic and rational integrals separately.
Combine the results and verify the answer by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 46 steps

Given —
\[I=\int\frac{3x+5}{x^3-x^2-x+1}\,dx\]
Factorise the denominator
Consider the denominator:
\[x^3-x^2-x+1\]
Grouping the terms:
\[\begin{aligned}x^3-x^2-x+1&=x^2(x-1)-1(x-1)\\ &=(x-1)(x^2-1)\end{aligned}\]
Using the difference of two squares:
\[x^2-1=(x-1)(x+1)\]
Therefore:
\[\boxed{x^3-x^2-x+1=(x-1)^2(x+1)}\]
Apply partial fraction decomposition.
Since (x - 1) is a repeated linear factor and (x + 1) is a distinct linear factor, assume:
\[\frac{3x+5}{(x-1)^2(x+1)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{(x-1)^2}\]
Taking the LCM on the right-hand side:
\[\begin{aligned}\frac{3x+5}{(x-1)^2(x+1)}=\frac{A(x-1)(x+1)+B(x-1)^2+C(x+1)}{(x-1)^2(x+1)}\end{aligned}\]
Since the denominators are identical, equating the numerators gives:
\[3x+5=A(x-1)(x+1)+B(x-1)^2+C(x+1)\]
Expand the expressions.
Using the algebraic identities:
\[(x-1)(x+1)=x^2-1\]

\[(x-1)^2=x^2-2x+1\]
Substituting these expressions:
\[3x+5=A(x^2-1)+B(x^2-2x+1)+C(x+1)\]
Expanding each term:
\[ \begin{aligned} 3x+5={}&Ax^2-A\\ &+Bx^2-2Bx+B\\ &+Cx+C \end{aligned} \]
Collect like terms
Grouping the coefficients of x², x and the constant terms:
\[3x+5=(A+B)x^2+(-2B+C)x+(-A+B+C)\]
Compare coefficients
Comparing the coefficients of x² on both sides:
\[A+B=0 \tag{1}\]
Comparing the coefficients of x:
\[-2B+C=3 \tag{2}\]
Comparing the constant terms:
\[-A+B+C=5 \tag{3}\]
Thus, we obtain the following three simultaneous equations:
\[ \begin{align} A+B&=0 \tag{1}\\ -2B+C&=3 \tag{2}\\ -A+B+C&=5 \tag{3} \end{align} \]
Determine the values of A, B and C
From equation (1):
\[A+B=0\]
Therefore:
\[\boxed{A=-B} \tag{4}\]
Substituting equation (4) into equation (3):
\[-A+B+C=5\]

\[-(-B)+B+C=5\]
Simplifying:
\[2B+C=5 \tag{5}\]
From equation (2):
\[-2B+C=3 \tag{6}\]
Subtracting equation (6) from equation (5):
\[ \begin{aligned} (2B+C)-(-2B+C)&=5-3\\ 2B+C+2B-C&=2\\ 4B&=2 \end{aligned} \]
Therefore:
\[\boxed{B=\frac12}\]
Using equation (4):
\[A=-B\]
Hence:
\[\boxed{A=-\frac12}\]
Substituting B = 1/2 into equation (6):
\[-2\left(\frac12\right)+C=3\]

\[-1+C=3\]
Therefore:
\[\boxed{C=4}\]
Thus, the required constants are:
\[\boxed{A=-\frac12,\quad B=\frac12,\quad C=4}\]
Write the partial fraction decomposition
Substituting the values of A, B and C:
\[\boxed{\frac{3x+5}{(x-1)^2(x+1)}=-\frac{1}{2(x-1)}+\frac{1}{2(x+1)}+\frac{4}{(x-1)^2}}\]
Substitute the decomposition into the integral
We have:
\[I=\int\frac{3x+5}{(x-1)^2(x+1)}\,dx\]
Using the partial fraction decomposition:
\[I=\int\left[-\frac{1}{2(x-1)}+\frac{1}{2(x+1)}+\frac{4}{(x-1)^2}\right]dx\]
Applying the linearity property of integration:
\[ \begin{aligned} I={}&-\frac12\int\frac{dx}{x-1}\\ &+\frac12\int\frac{dx}{x+1}\\ &+4\int\frac{dx}{(x-1)^2} \end{aligned} \]
Evaluate the first integral
Consider:
\[-\frac12\int\frac{dx}{x-1}\]
Using the standard logarithmic integration formula:
\[\int\frac{dx}{x-a}=\log|x-a|+C\]
We obtain:
\[\boxed{-\frac12\int\frac{dx}{x-1}=-\frac12\log|x-1|}\]
Evaluate the second integral
Consider:
\[\frac12\int\frac{dx}{x+1}\]
Using logarithmic integration:
\[\boxed{\frac12\int\frac{dx}{x+1}=\frac12\log|x+1|}\]
Evaluate the third integral
Consider:
\[4\int\frac{dx}{(x-1)^2}\]
Using substitution, let:
\[t=x-1\]
Differentiating both sides:
\[dt=dx\]
Substituting into the integral:
\[4\int\frac{dt}{t^2}\]
Expressing the negative power:
\[4\int t^{-2}\,dt\]
Applying the power rule of integration:
\[ \begin{aligned} 4\int t^{-2}\,dt &= 4\left(\frac{t^{-2+1}}{-2+1}\right)\\ &= 4\left(\frac{t^{-1}}{-1}\right)\\ &=-4t^{-1}\\ &=-\frac4t \end{aligned} \]
Substituting t = x - 1:
\[\boxed{4\int\frac{dx}{(x-1)^2}=-\frac4{x-1}}\]
Combine all the integrals
\[ \begin{aligned} I={}&-\frac12\log|x-1|\\ &+\frac12\log|x+1|\\ &-\frac4{x-1}+C \end{aligned} \]
Using the logarithmic identity:
\[\log a-\log b=\log\left(\frac ab\right)\]
We obtain:
\[I=\frac12\log\left|\frac{x+1}{x-1}\right|-\frac4{x-1}+C\]
Therefore, the final answer is:
\[\boxed{\displaystyle\int\frac{3x+5}{x^3-x^2-x+1}\,dx=\frac12\log\left|\frac{x+1}{x-1}\right|-\frac4{x-1}+C}\]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int\frac{3x+5}{x^3-x^2-x+1}\,dx=\frac12\log\left|\frac{x+1}{x-1}\right|-\frac4{x-1}+C}\]

🎯 Exam Significance

Exam Significance

›

This problem is useful for CBSE Class 12 Mathematics board examination preparation because it combines polynomial factorisation, repeated linear factors and partial fraction integration.

Factorisation of cubic polynomials: It tests the ability to recognise and factorise algebraic expressions using grouping and the difference of squares.
Repeated linear factors: It reinforces the correct structure of partial fraction decomposition when a linear factor occurs more than once.
Coefficient comparison: Students practise forming and solving simultaneous equations to determine unknown constants.
Logarithmic integration: It strengthens the application of standard logarithmic integration formulae.
Power rule: It tests the integration of negative powers and the importance of maintaining the correct sign.
Stepwise mathematical presentation: The solution demonstrates a systematic approach suitable for descriptive examination questions.

Significance for Competitive Entrance Examinations

The techniques used in this question are relevant to JEE Main, JEE Advanced, BITSAT and other engineering entrance examinations.

Polynomial factorisation: Recognising algebraic patterns helps simplify complicated rational functions.
Repeated-factor decomposition: Understanding repeated linear factors is essential for handling more complex rational integrals.
Efficient calculation: Substitution of suitable values into the numerator identity can reduce the work involved in finding constants.
Algebraic precision: Correct handling of signs and fractions is essential in objective-type questions.
Integration techniques: The problem combines logarithmic integration with the power rule, two fundamental integration methods.
Verification skills: Differentiating an antiderivative provides a reliable way to check an answer.

🚫 Common Mistakes

Common Mistakes

›

Incomplete factorisation: The denominator must be factorised completely as (x - 1)²(x + 1).
Incorrect partial fraction structure: A repeated factor (x - 1)² requires both A/(x - 1) and C/(x - 1)².
Using an incorrect denominator: Do not write C/[(x - 1)(x + 1)] as the repeated-factor term.
Incorrect expansion: Remember that (x - 1)(x + 1) = x² - 1.
Incorrect coefficient comparison: The correct equations are A + B = 0, -2B + C = 3 and -A + B + C = 5.
Sign errors: Carefully substitute A = -B into the constant-term equation.
Incorrect integration of repeated factors: The integral of 1/(x - 1)² is -1/(x - 1), not +1/(x - 1).
Incorrect logarithmic simplification: Ensure that the quotient inside the logarithm is arranged consistently with the signs of the original terms.
Omitting the integration constant: Always include C in the final indefinite integral.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 8 points

Always factorise the denominator completely before attempting partial fraction decomposition.
A repeated linear factor requires a separate partial fraction for every power of that factor.
The correct decomposition for this problem contains three terms: A/(x - 1), B/(x + 1) and C/(x - 1)².
Coefficient comparison and direct substitution are two useful methods for determining unknown constants.
Simple linear factors produce logarithmic terms upon integration.
Repeated linear factors with power two produce rational terms after integration.
The negative sign in the integration of (x - 1)⁻² must be handled carefully.
Differentiation provides an effective method for verifying the final antiderivative.

📘 Concept & Theory

Theory and Concept

›

Integration of Rational Functions with Distinct Linear Factors

Partial fraction decomposition is an important technique for integrating rational functions. It involves expressing a complicated rational expression as a sum of simpler fractions whose integrals can be evaluated using standard integration formulae.

The first step is to factorise the denominator completely.

In the given problem, the denominator is:

\[ (x^2-1)(2x+3) \]

Using the difference of two squares:

\[ x^2-1=(x-1)(x+1) \]

Therefore:

\[\boxed{(x^2-1)(2x+3)=(x-1)(x+1)(2x+3)}\]

The denominator contains three distinct linear factors:

(x - 1)
(x + 1)
(2x + 3)

For a rational function whose denominator contains three distinct linear factors, the appropriate partial fraction decomposition is:

\[\frac{P(x)}{(x-a)(x-b)(cx+d)}=\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{cx+d}\]

Thus, for the given problem:

\[\boxed{\frac{2x-3}{(x-1)(x+1)(2x+3)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{2x+3}}\]

Here, A, B and C are constants to be determined.

Important integration formulae:

For a linear expression with coefficient 1:

\[\int\frac{dx}{x-a}=\log|x-a|+C\]

For a linear expression with coefficient a:

\[\boxed{\int\frac{dx}{ax+b}=\frac1a\log|ax+b|+C}\]

The coefficient of x in the denominator must be accounted for during integration. This is particularly important when integrating expressions such as 1/(2x + 3).

🗺️ Solution Roadmap

Step-by-step Plan

›

Factorise x² - 1 using the difference of two squares.
Write the appropriate partial fraction decomposition for the three distinct linear factors.
Multiply both sides by the common denominator.
Expand the expressions and collect the coefficients of x², x and the constant term.
Form three simultaneous equations and calculate A, B and C.
Substitute the constants into the original integral.
Evaluate each integral separately using logarithmic integration.
Apply the correct coefficient adjustment when integrating 1/(2x + 3).
Combine the results and verify the answer by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 42 steps

Given —
\[I=\int\frac{2x-3}{(x^2-1)(2x+3)}\,dx\]
Factorise the denominator
Using the algebraic identity for the difference of two squares:
\[x^2-1=(x-1)(x+1)\]
Therefore:
\[\begin{aligned}I&=\int\frac{2x-3}{(x-1)(x+1)(2x+3)}\,dx\end{aligned}\]
Apply partial fraction decomposition
Since the denominator contains three distinct linear factors, assume:
\[\frac{2x-3}{(x-1)(x+1)(2x+3)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{2x+3}\]
Taking the LCM on the right-hand side:
\[\begin{aligned}\frac{2x-3}{(x-1)(x+1)(2x+3)}=\frac{A(x+1)(2x+3)+B(x-1)(2x+3)+C(x-1)(x+1)}{(x-1)(x+1)(2x+3)}\end{aligned}\]
Since the denominators on both sides are identical, equating the numerators gives:
\[\boxed{2x-3=A(x+1)(2x+3)+B(x-1)(2x+3)+C(x-1)(x+1)}\]
Expand the products
First, expand the product containing A:
\[ \begin{aligned} A(x+1)(2x+3) &=A(2x^2+3x+2x+3)\\ &=A(2x^2+5x+3)\\ &=2Ax^2+5Ax+3A \end{aligned} \]
Next, expand the product containing B:
\[ \begin{aligned} B(x-1)(2x+3) &=B(2x^2+3x-2x-3)\\ &=B(2x^2+x-3)\\ &=2Bx^2+Bx-3B \end{aligned} \]
Finally, expand the product containing C:
\[ \begin{aligned} C(x-1)(x+1) &=C(x^2-1)\\ &=Cx^2-C \end{aligned} \]
Substituting all the expanded expressions into the numerator identity:
\[ \begin{aligned} 2x-3={}&2Ax^2+5Ax+3A\\ &+2Bx^2+Bx-3B\\ &+Cx^2-C \end{aligned} \]
Collect like terms
\[2x-3=(2A+2B+C)x^2+(5A+B)x+(3A-3B-C)\]
Compare coefficients
Comparing the coefficients of x² on both sides:
\[2A+2B+C=0 \tag{1}\]
Comparing the coefficients of x:
\[5A+B=2 \tag{2}\]
Comparing the constant terms:
\[3A-3B-C=-3 \tag{3}\]
Therefore, we obtain the following three simultaneous equations:
\[ \begin{align} 2A+2B+C&=0 \tag{1}\\ 5A+B&=2 \tag{2}\\ 3A-3B-C&=-3 \tag{3} \end{align} \]
Determine the values of A, B and C.
From equation (1):
\[2A+2B+C=0\]
Therefore:
\[C=-2A-2B \tag{4}\]
Substituting equation (4) into equation (3):
\[3A-3B-(-2A-2B)=-3\]
Removing the brackets:
\[3A-3B+2A+2B=-3\]
Combining like terms:
\[5A-B=-3 \tag{5}\]
From equation (2):
\[5A+B=2 \tag{6}\]
Adding equations (5) and (6):
\[ \begin{aligned} (5A-B)+(5A+B)&=-3+2\\ 10A&=-1 \end{aligned} \]
Therefore:
\[\boxed{A=-\frac1{10}}\]
Substituting A = -1/10 into equation (6):
\[5\left(-\frac1{10}\right)+B=2\]

\[-\frac12+B=2\]
Therefore:
\[\boxed{B=\frac52}\]
Now, substituting the values of A and B into equation (4):
\[C=-2\left(-\frac1{10}\right)-2\left(\frac52\right)\]

\[C=\frac15-5\]
Therefore:
\[\boxed{C=-\frac{24}{5}}\]
Thus, the correct values of the constants are:
\[\boxed{A=-\frac1{10},\quad B=\frac52,\quad C=-\frac{24}{5}}\]
Write the partial fraction decomposition.
Substituting the values of A, B and C:
\[\boxed{\frac{2x-3}{(x-1)(x+1)(2x+3)}=-\frac{1}{10(x-1)}+\frac{5}{2(x+1)}-\frac{24}{5(2x+3)}}\]
Substitute the decomposition into the integral.
We have:
\[I=\int\frac{2x-3}{(x^2-1)(2x+3)}\,dx\]
Using the partial fraction decomposition:
\[I=\int\left[-\frac{1}{10(x-1)}+\frac{5}{2(x+1)}-\frac{24}{5(2x+3)}\right]dx\]
Applying the linearity property of integration:
\[ \begin{aligned} I={}&-\frac1{10}\int\frac{dx}{x-1}\\ &+\frac52\int\frac{dx}{x+1}\\ &-\frac{24}{5}\int\frac{dx}{2x+3} \end{aligned} \]
Evaluate the first integral
Using the standard logarithmic integration formula:
\[\int\frac{dx}{x-a}=\log|x-a|+C\]
We obtain:
\[\boxed{-\frac1{10}\int\frac{dx}{x-1}=-\frac1{10}\log|x-1|}\]
Evaluate the second integral
Applying logarithmic integration:
\[\boxed{\frac52\int\frac{dx}{x+1}=\frac52\log|x+1|}\]
Evaluate the third integral using substitution
Consider:
\[-\frac{24}{5}\int\frac{dx}{2x+3}\]
Let:
\[t=2x+3\]
Differentiating both sides:
\[dt=2dx\]
Therefore:
\[dx=\frac{dt}{2}\]
Substituting into the integral:
\[-\frac{24}{5}\int\frac{1}{t}\cdot\frac{dt}{2}\]
Simplifying:
\[-\frac{12}{5}\int\frac{dt}{t}\]
Using the logarithmic integration formula:
\[-\frac{12}{5}\log|t|+C\]
Substituting t = 2x + 3:
\[\boxed{-\frac{24}{5}\int\frac{dx}{2x+3}=-\frac{12}{5}\log|2x+3|}\]
Combine all the integrals
\[ \begin{aligned} I={}&-\frac1{10}\log|x-1|\\ &+\frac52\log|x+1|\\ &-\frac{12}{5}\log|2x+3|+C \end{aligned} \]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int\frac{2x-3}{(x^2-1)(2x+3)}\,dx=-\frac1{10}\log|x-1|+\frac52\log|x+1|-\frac{12}{5}\log|2x+3|+C}\]

🎯 Exam Significance

Exam Significance

›

This problem develops several important techniques for CBSE Class 12 Mathematics:

Factorisation: Applying the difference of two squares to simplify the denominator.
Partial fraction decomposition: Expressing a rational function with distinct linear factors as simpler fractions.
Coefficient comparison: Forming simultaneous equations from an algebraic identity.
Logarithmic integration: Applying standard formulae to linear denominators.
Substitution: Recognising the coefficient adjustment required for denominators such as 2x + 3.
Verification: Checking an antiderivative by differentiation.

Significance for Competitive Entrance Examinations

The underlying techniques are useful for JEE Main, JEE Advanced, BITSAT and other engineering entrance examinations.

Recognising factorisation patterns in polynomial denominators.
Constructing the correct partial fraction decomposition.
Solving simultaneous equations efficiently.
Applying substitution correctly to logarithmic integrals.
Detecting incorrect answer options through differentiation.
Maintaining algebraic accuracy in multi-step integration problems.

🚫 Common Mistakes

Common Mistakes

›

Failing to factorise x² - 1 as (x - 1)(x + 1).
Using an incorrect partial fraction structure.
Expanding (x + 1)(2x + 3) incorrectly.
Forgetting the coefficient 2 in the derivative of 2x + 3.
Using incorrect coefficients when comparing the constant terms.
Combining logarithms before checking the signs of their coefficients.
Failing to verify the final answer by differentiation.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

Factorise the denominator completely before applying partial fractions.
Three distinct linear factors require three corresponding partial fraction terms.
Use coefficient comparison carefully and check all three equations.
The integral of 1/(ax + b) includes the factor 1/a.
Differentiate the final expression to confirm that it reproduces the original integrand.

📘 Concept & Theory

Theory and Concept

›

Integration of Rational Functions Using Distinct Linear Factors

Partial fraction decomposition is a technique used to express a complicated rational function as a sum of simpler rational expressions. These simpler expressions can then be integrated using standard integration formulae.

The first step is to factorise the denominator completely.

In the given integral, the denominator is:

\[(x+1)(x^2-4)\]

Using the difference of two squares:

\[x^2-4=(x-2)(x+2)\]

Therefore:

\[\boxed{(x+1)(x^2-4)=(x+1)(x-2)(x+2)}\]

The denominator contains three distinct linear factors:

(x + 1)
(x - 2)
(x + 2)

For a rational function whose denominator contains three distinct linear factors, the appropriate partial fraction decomposition is:

\[\frac{P(x)}{(x-a)(x-b)(x-c)}=\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{x-c}\]

Thus, for the given problem, we assume:

\[\boxed{\frac{5x}{(x+1)(x-2)(x+2)}=\frac{A}{x+1}+\frac{B}{x-2}+\frac{C}{x+2}}\]

Here, A, B and C are constants to be determined.

Important integration formula:

\[\boxed{\int\frac{dx}{x-a}=\log|x-a|+C}\]

Since the linear expressions in this question have coefficient 1 for x, each partial fraction can be integrated directly using the logarithmic integration formula.

🗺️ Solution Roadmap

Step-by-step Plan

›

Factorise the quadratic expression x² - 4.
Write the appropriate partial fraction decomposition for the three distinct linear factors.
Multiply both sides by the common denominator to eliminate fractions.
Expand the products and collect the coefficients of x², x and the constant term.
Form three simultaneous equations and determine A, B and C.
Substitute the constants into the original integral.
Evaluate each integral separately using logarithmic integration.
Combine the logarithmic terms and simplify the final expression.
Verify the result by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 46 steps

Given —
\[I=\int\frac{5x}{(x+1)(x^2-4)}\,dx\]
Factorise the denominator
Using the difference of two squares:
\[x^2-4=(x-2)(x+2)\]
Substituting this factorisation into the integral:
\[I=\int\frac{5x}{(x+1)(x-2)(x+2)}\,dx\]
Apply partial fraction decomposition
Since the denominator contains three distinct linear factors, assume:
\[\frac{5x}{(x+1)(x-2)(x+2)}=\frac{A}{x+1}+\frac{B}{x-2}+\frac{C}{x+2}\]
Taking the LCM on the right-hand side:
\[\begin{aligned}\frac{5x}{(x+1)(x-2)(x+2)}=\frac{A(x-2)(x+2)+B(x+1)(x+2)+C(x+1)(x-2)}{(x+1)(x-2)(x+2)}\end{aligned}\]
Since the denominators on both sides are identical, equating the numerators gives:
\[\boxed{5x=A(x-2)(x+2)+B(x+1)(x+2)+C(x+1)(x-2)}\]
Expand the products
First, expand the product containing A:
\[\begin{aligned}A(x-2)(x+2)&=A(x^2-4)\\ &=Ax^2-4A\end{aligned}\]
Next, expand the product containing B:
\[\begin{aligned}B(x+1)(x+2)&=B(x^2+2x+x+2)\\ &=B(x^2+3x+2)\\ &=Bx^2+3Bx+2B\end{aligned}\]
Finally, expand the product containing C:
\[\begin{aligned}C(x+1)(x-2)&=C(x^2-2x+x-2)\\ &=C(x^2-x-2)\\ &=Cx^2-Cx-2C\end{aligned}\]
Substituting all the expanded expressions into the numerator identity:
\[\begin{aligned}5x={}&Ax^2-4A\\ &+Bx^2+3Bx+2B\\ &+Cx^2-Cx-2C\end{aligned}\]
Collect like terms
Grouping the coefficients of x², x and the constant terms:
\[5x=(A+B+C)x^2+(3B-C)x+(-4A+2B-2C)\]
Compare coefficients
Comparing the coefficients of x² on both sides:
\[A+B+C=0 \tag{1}\]
Comparing the coefficients of x:
\[3B-C=5 \tag{2}\]
Comparing the constant terms:
\[-4A+2B-2C=0 \tag{3}\]
Thus, the three simultaneous equations are:
\[ \begin{align} A+B+C&=0 \tag{1}\\ 3B-C&=5 \tag{2}\\ -4A+2B-2C&=0 \tag{3} \end{align} \]
Determine the values of A, B and C.
From equation (2):
\[3B-C=5\]
Therefore:
\[\boxed{C=3B-5} \tag{4}\]
Substituting equation (4) into equation (1):
\[A+B+(3B-5)=0\]
Simplifying:
\[A+4B-5=0\]
Therefore:
\[\boxed{A=5-4B} \tag{5}\]
Now, divide equation (3) by 2:
\[-2A+B-C=0\]
Therefore:
\[C=B-2A \tag{6}\]
Equating equations (4) and (6):
\[3B-5=B-2A\]
Rearranging:
\[2B+2A=5\]
Therefore:
\[\boxed{A+B=\frac52} \tag{7}\]
Using equation (5), substitute A = 5 - 4B into equation (7):
\[(5-4B)+B=\frac52\]
Simplifying:
\[5-3B=\frac52\]
Transposing:
\[-3B=\frac52-5\]

\[-3B=-\frac52\]
Dividing both sides by -3:
\[\boxed{B=\frac56}\]
Substituting B = 5/6 into equation (5):
\[A=5-4\left(\frac56\right)\]

\[A=5-\frac{20}{6}\]
Taking the LCM:
\[A=\frac{30-20}{6}\]
Therefore:
\[\boxed{A=\frac53}\]
Finally, substituting B = 5/6 into equation (4):
\[C=3\left(\frac56\right)-5\]

\[C=\frac{15}{6}-5\]

\[C=\frac52-5\]
Therefore:
\[\boxed{C=-\frac52}\]
Thus, the required constants are:
\[\boxed{A=\frac53,\quad B=\frac56,\quad C=-\frac52}\]
Write the partial fraction decomposition
Substituting the values of A, B and C:
\[\boxed{\frac{5x}{(x+1)(x-2)(x+2)}=\frac{5}{3(x+1)}+\frac{5}{6(x-2)}-\frac{5}{2(x+2)}}\]
Substitute the decomposition into the integral
We have:
\[I=\int\frac{5x}{(x+1)(x^2-4)}\,dx\]
Using the partial fraction decomposition:
\[I=\int\left[\frac{5}{3(x+1)}+\frac{5}{6(x-2)}-\frac{5}{2(x+2)}\right]dx\]
Applying the linearity property of integration:
\[\begin{aligned}I={}&\frac53\int\frac{dx}{x+1}\\ &+\frac56\int\frac{dx}{x-2}\\ &-\frac52\int\frac{dx}{x+2}\end{aligned}\]
Evaluate the first integral
Using the logarithmic integration formula:
\[\int\frac{dx}{x-a}=\log|x-a|+C\]
We obtain:
\[\boxed{\frac53\int\frac{dx}{x+1}=\frac53\log|x+1|}\]
Evaluate the second integral
Applying logarithmic integration:
\[\boxed{\frac56\int\frac{dx}{x-2}=\frac56\log|x-2|}\]
Evaluate the third integral
Using the logarithmic integration formula:
\[\boxed{-\frac52\int\frac{dx}{x+2}=-\frac52\log|x+2|}\]
Combine all the integrals
\[ \begin{aligned} I={}& \frac53\log|x+1|\\ &+\frac56\log|x-2|\\ &-\frac52\log|x+2|+C \end{aligned} \]
Using the logarithmic identity:
\[a\log m+b\log n=\log(m^a n^b)\]
We can express the result in a single logarithm:
\[I=\log\left|\frac{(x+1)^{5/3}(x-2)^{5/6}}{(x+2)^{5/2}}\right|+C\]
However, the separate logarithmic expression is generally clearer for real-valued integration because it avoids ambiguity associated with fractional powers of negative quantities.
Therefore,
\[\boxed{\begin{aligned}\displaystyle\int\frac{5x}{(x+1)(x^2-4)}\,dx={}&\frac53\log|x+1|\\ &+\frac56\log|x-2|\\&-\frac52\log|x+2|+C\end{aligned}}\]

💡 Answer

Final Answer

›

\[\boxed{\begin{aligned}\displaystyle\int\frac{5x}{(x+1)(x^2-4)}\,dx={}&\frac53\log|x+1|\\ &+\frac56\log|x-2|\\&-\frac52\log|x+2|+C\end{aligned}}\]

🎯 Exam Significance

Exam Significance

›

This problem is useful for CBSE Class 12 Mathematics board examination preparation because it reinforces the application of partial fractions to rational functions with distinct linear factors.

Factorisation: It tests the application of the difference of two squares to factorise a quadratic expression.
Partial fraction decomposition: Students learn to express a rational function with three distinct linear factors as three simpler fractions.
Coefficient comparison: It develops the ability to form and solve simultaneous linear equations.
Logarithmic integration: It reinforces the standard formula for integrating the reciprocal of a linear expression.
Algebraic accuracy: It highlights the importance of maintaining correct signs while expanding and simplifying expressions.
Verification: Differentiation provides a method to confirm the correctness of the final antiderivative.

Significance for Competitive Entrance Examinations

The techniques used in this problem are relevant to JEE Main, JEE Advanced, BITSAT and other engineering entrance examinations.

Recognition of algebraic identities: Quickly identifying the difference of two squares helps simplify rational functions.
Partial fraction techniques: Understanding distinct linear factors is essential for more complicated integration problems.
Efficient calculation: Strategic substitution can simplify the determination of unknown constants.
Algebraic precision: Correct expansion and coefficient comparison reduce calculation errors.
Logarithmic properties: Combining logarithmic expressions helps simplify equivalent mathematical answers.
Answer verification: Differentiation can be used to check the validity of an answer in subjective and objective problems.

🚫 Common Mistakes

Common Mistakes

›

Incomplete factorisation: Remember that x² - 4 = (x - 2)(x + 2).
Incorrect partial fraction structure: Use one fraction for each distinct linear factor.
Incorrect expansion: Carefully expand (x + 1)(x + 2) and (x + 1)(x - 2).
Incorrect coefficient comparison: The correct equations are A + B + C = 0, 3B - C = 5 and -4A + 2B - 2C = 0.
Sign errors: Pay particular attention to the negative sign in the expression C(x + 1)(x - 2).
Incorrect constants: Verify all three constants against the original numerator identity.
Incorrect logarithmic signs: Preserve the signs of the coefficients when integrating each partial fraction.
Missing integration constant: Always include C in the final indefinite integral.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

Factorise the denominator completely before applying partial fraction decomposition.
The difference of two squares identity is useful for factorising expressions such as x² - 4.
Three distinct linear factors require three corresponding partial fraction terms.
Coefficient comparison and direct substitution are both valid methods for finding unknown constants.
Each simple linear denominator produces a logarithmic term upon integration.
Check the signs of all constants before integrating.
Differentiation is an effective method of verifying an indefinite integral.

📘 Concept & Theory

Theory and Concept

›

Integration of Improper Rational Functions

A rational function is an algebraic expression represented as the quotient of two polynomials. Depending on the degrees of the numerator and denominator, rational functions are classified into two categories:

Proper rational function: The degree of the numerator is less than the degree of the denominator.
Improper rational function: The degree of the numerator is greater than or equal to the degree of the denominator.

In the given problem:

\[\frac{x^3+x+1}{x^2-1}\]

The degree of the numerator is 3, whereas the degree of the denominator is 2.

Therefore:

\[3>2\]

Hence, the given rational function is an improper rational function.

Important rule:

Whenever the degree of the numerator is greater than or equal to the degree of the denominator, polynomial division must be performed before applying partial fraction decomposition.

Polynomial division expresses the improper rational function as the sum of a polynomial and a proper rational function.

For example:

\[\frac{P(x)}{Q(x)}=S(x)+\frac{R(x)}{Q(x)}\]

where:

S(x) is the polynomial quotient.
R(x) is the polynomial remainder.
The degree of R(x) is less than the degree of Q(x).

After polynomial division, the proper rational function can be decomposed into partial fractions.

Important integration formulae:

Power rule of integration:

\[\boxed{\int x^n\,dx=\frac{x^{n+1}}{n+1}+C,\quad n\neq-1}\]

Logarithmic integration:

\[\boxed{\int\frac{dx}{x-a}=\log|x-a|+C}\]

These formulae will be used to integrate the polynomial quotient and the partial fractions separately.

🗺️ Solution Roadmap

Step-by-step Plan

›

Compare the degrees of the numerator and denominator to identify the rational function as improper.
Perform polynomial division to express the rational function as a polynomial plus a proper rational function.
Factorise the denominator of the proper rational function using the difference of two squares.
Assume the appropriate partial fraction decomposition.
Multiply both sides by the common denominator and compare coefficients.
Determine the unknown constants A and B.
Substitute the constants into the original integral.
Integrate the polynomial and the two partial fractions separately.
Combine the results and verify the final answer by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 38 steps

Given —
\[I=\int\frac{x^3+x+1}{x^2-1}\,dx\]
Identify the nature of the rational function.
The degree of the numerator is 3, and the degree of the denominator is 2.

Since the degree of the numerator is greater than the degree of the denominator, the given rational function is improper.

Therefore, polynomial division must be performed before applying partial fraction decomposition.
Perform polynomial division
Divide the numerator x³ + x + 1 by the denominator x² - 1.
The first term of the quotient is obtained by dividing the highest-degree terms:
\[\frac{x^3}{x^2}=x\]
Therefore, the quotient begins with x
Multiplying the divisor by x:
\[x(x^2-1)=x^3-x\]
Subtracting this expression from the numerator:
\[ \begin{aligned} &(x^3+x+1)-(x^3-x)\\ &=x^3+x+1-x^3+x\\ &=2x+1 \end{aligned} \]
Thus, the quotient is x and the remainder is 2x + 1
Using the polynomial division identity:
\[\text{Dividend}=\text{Divisor}\times\text{Quotient}+\text{Remainder}\]
We obtain:
\[x^3+x+1=x(x^2-1)+(2x+1)\]
Dividing both sides by x² - 1:
\[\boxed{\frac{x^3+x+1}{x^2-1}=x+\frac{2x+1}{x^2-1}}\]
Factorise the denominator of the proper rational function.
Consider the denominator:
\[x^2-1\]
Using the difference of two squares identity:
\[a^2-b^2=(a-b)(a+b)\]
We obtain:
\[\boxed{x^2-1=(x-1)(x+1)}\]
Therefore, the proper rational function becomes:
\[\frac{2x+1}{(x-1)(x+1)}\]
Apply partial fraction decomposition
Since the denominator contains two distinct linear factors, assume:
\[\frac{2x+1}{(x-1)(x+1)}=\frac{A}{x-1}+\frac{B}{x+1}\]
Taking the LCM on the right-hand side:
\[\frac{2x+1}{(x-1)(x+1)}=\frac{A(x+1)+B(x-1)}{(x-1)(x+1)}\]
Since the denominators on both sides are identical, equating the numerators gives:
\[\boxed{2x+1=A(x+1)+B(x-1)}\]
Expand the expressions
Expanding the products:
\[2x+1=Ax+A+Bx-B\]
Collecting like terms:
\[2x+1=(A+B)x+(A-B)\]
Compare coefficients
Comparing the coefficients of x on both sides:
\[A+B=2 \tag{1}\]
Comparing the constant terms:
\[A-B=1 \tag{2}\]
Thus, we obtain two simultaneous equations:
\[\begin{align}A+B&=2 \tag{1}\\ A-B&=1 \tag{2}\end{align}\]
Determine the values of A and B.
Adding equations (1) and (2):
\[\begin{aligned}(A+B)+(A-B)&=2+1\\ A+B+A-B&=3\\ 2A&=3\end{aligned}\]
Dividing both sides by 2:
\[\boxed{A=\frac32}\]
Substituting A = 3/2 into equation (1):
\[\frac32+B=2\]
Transposing:
\[B=2-\frac32\]
Therefore:
\[\boxed{B=\frac12}\]
Thus, the required constants are:
\[\boxed{A=\frac32,\qquad B=\frac12}\]
Write the partial fraction decomposition
Substituting the values of A and B:
\[\boxed{\frac{2x+1}{x^2-1}=\frac{3}{2(x-1)}+\frac{1}{2(x+1)}}\]
Combining this with the polynomial division result:
\[\boxed{\frac{x^3+x+1}{x^2-1}=x+\frac{3}{2(x-1)}+\frac{1}{2(x+1)}}\]
Substitute the decomposition into the integral
We have:
\[I=\int\frac{x^3+x+1}{x^2-1}\,dx\]
Substituting the simplified expression:
\[I=\int\left[x+\frac{3}{2(x-1)}+\frac{1}{2(x+1)}\right]dx\]
Applying the linearity property of integration:
\[ \begin{aligned} I={}&\int x\,dx\\ &+\frac32\int\frac{dx}{x-1}\\ &+\frac12\int\frac{dx}{x+1} \end{aligned} \]
Evaluate the first integral.
Consider:
\[\int x\,dx\]
Using the power rule of integration:
\[\int x^n\,dx=\frac{x^{n+1}}{n+1}+C\]
Putting n = 1:
\[\begin{aligned}\int x\,dx&=\frac{x^{1+1}}{1+1}\\ &=\frac{x^2}{2}\end{aligned}\]
Therefore:
\[\boxed{\int x\,dx=\frac{x^2}{2}}\]
Evaluate the second integral
Consider:
\[\frac32\int\frac{dx}{x-1}\]
Using the standard logarithmic integration formula:
\[\int\frac{dx}{x-a}=\log|x-a|+C\]
We obtain:
\[\boxed{\frac32\int\frac{dx}{x-1}=\frac32\log|x-1|}\]
Evaluate the third integral
Consider:
\[\frac12\int\frac{dx}{x+1}\]
Applying the logarithmic integration formula:
\[\boxed{\frac12\int\frac{dx}{x+1}=\frac12\log|x+1|}\]
Combine all the integrals
\[ \begin{aligned} I={}&\frac{x^2}{2}\\ &+\frac32\log|x-1|\\ &+\frac12\log|x+1|+C \end{aligned} \]

💡 Answer

Final Answer

›

\[ \boxed{ \begin{aligned} \displaystyle \int\frac{x^3+x+1}{x^2-1}\,dx ={}& \frac{x^2}{2}\\ &+\frac32\log|x-1|\\ &+\frac12\log|x+1|+C \end{aligned} } \]

🎯 Exam Significance

Exam Significance

›

This problem is important for CBSE Class 12 Mathematics because it combines polynomial division, partial fraction decomposition and standard integration techniques.

Improper rational functions: It tests the ability to identify when polynomial division is necessary before integration.
Polynomial division: Students practise obtaining the quotient and remainder of a polynomial division.
Factorisation: It reinforces the use of the difference of two squares identity.
Partial fraction decomposition: Students learn to decompose a proper rational function with distinct linear factors.
Simultaneous equations: It develops accuracy in determining unknown constants through coefficient comparison.
Integration formulae: It combines the power rule with logarithmic integration.
Stepwise presentation: The solution demonstrates how to organise a multi-stage integration problem clearly in a descriptive examination.

Significance for Competitive Entrance Examinations

The techniques used in this problem are useful for JEE Main, JEE Advanced, BITSAT and other engineering entrance examinations.

Degree comparison: Recognising an improper rational function helps determine the correct starting method.
Polynomial division: Efficient division simplifies complicated rational expressions.
Algebraic identities: Quick factorisation reduces unnecessary calculations.
Partial fractions: Decomposition is a fundamental technique for evaluating rational integrals.
Logarithmic integration: Students strengthen their understanding of integrals involving reciprocal linear expressions.
Mixed integration techniques: The question combines polynomial integration and logarithmic integration in one problem.
Verification skills: Differentiation helps detect incorrect signs, coefficients and missing terms in an antiderivative.

🚫 Common Mistakes

Common Mistakes

›

Skipping polynomial division: Since the degree of the numerator is greater than that of the denominator, polynomial division must be performed first.
Incorrect quotient: The quotient is x, and the remainder is 2x + 1.
Incorrect subtraction: Remember that (x³ + x + 1) - (x³ - x) = 2x + 1.
Incomplete factorisation: Always factorise x² - 1 as (x - 1)(x + 1).
Incorrect partial fraction structure: Since the denominator contains two distinct linear factors, use A/(x - 1) + B/(x + 1).
Coefficient comparison errors: The correct equations are A + B = 2 and A - B = 1.
Incorrect logarithmic coefficients: The constants 3/2 and 1/2 must be retained when integrating the partial fractions.
Forgetting the polynomial term: Do not omit the integral of x obtained through polynomial division.
Missing integration constant: Always include C in the final indefinite integral.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

Always compare the degrees of the numerator and denominator before applying partial fractions.
Improper rational functions must first be simplified using polynomial division.
Polynomial division expresses an improper rational function as a polynomial plus a proper rational function.
Factorise the denominator completely before decomposing the proper rational part.
Distinct linear factors require separate partial fraction terms.
Integrate polynomial terms using the power rule and reciprocal linear terms using logarithmic integration.
Verify the final antiderivative by differentiation.

📘 Concept & Theory

Theory and Concept

›

Integration of Rational Functions Containing an Irreducible Quadratic Factor

Partial fraction decomposition is an important technique for integrating rational functions. It expresses a complicated rational expression as a sum of simpler fractions that can be integrated independently.

In the given problem, the denominator contains two factors:

A linear factor: (1 - x).
An irreducible quadratic factor: (1 + x²).

The quadratic expression 1 + x² cannot be factorised into real linear factors because the equation x² + 1 = 0 has no real roots.

Rule for partial fraction decomposition:

When a rational function has a distinct linear factor and an irreducible quadratic factor in its denominator, the appropriate decomposition is:

\[\frac{P(x)}{(ax+b)(cx^2+dx+e)} = \frac{A}{ax+b} + \frac{Bx+C}{cx^2+dx+e}\]

Notice that the numerator over the irreducible quadratic factor must be a linear expression, Bx + C, rather than just a constant.

Therefore, for the given problem, we assume:

\[\boxed{\frac{2}{(1-x)(1+x^2)}=\frac{A}{1-x}+\frac{Bx+C}{1+x^2}}\]

Important integration formulae:

Integration of a linear expression:

\[\boxed{\int\frac{dx}{a-x}=-\log|a-x|+C}\]

Integration using substitution:

\[\boxed{\int\frac{x}{x^2+a^2}\,dx=\frac12\log(x^2+a^2)+C}\]

Integration of an irreducible quadratic expression:

\[\boxed{\int\frac{dx}{1+x^2}=\tan^{-1}x+C}\]

These formulae will be used to evaluate the three resulting terms independently.

🗺️ Solution Roadmap

Step-by-step Plan

›

Identify the linear and irreducible quadratic factors in the denominator.
Assume the appropriate partial fraction decomposition with three unknown constants A, B and C.
Multiply both sides by the common denominator.
Expand the expressions and collect the coefficients of x², x and the constant term.
Form three simultaneous equations and determine A, B and C.
Substitute the constants into the original integral.
Separate the resulting expression into three simpler integrals.
Evaluate the linear denominator using substitution, the quadratic numerator term using logarithmic integration, and the remaining quadratic term using inverse trigonometric integration.
Combine the results and verify the final answer by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 45 steps

Given —
\[I=\int\frac{2}{(1-x)(1+x^2)}\,dx\]
Apply partial fraction decomposition
Since the denominator contains one linear factor and one irreducible quadratic factor, assume:
\[\frac{2}{(1-x)(1+x^2)}=\frac{A}{1-x}+\frac{Bx+C}{1+x^2}\]
Taking the LCM on the right-hand side:
\[\frac{2}{(1-x)(1+x^2)}=\frac{A(1+x^2)+(Bx+C)(1-x)}{(1-x)(1+x^2)}\]
Since the denominators on both sides are identical, equating the numerators gives:
\[\boxed{2=A(1+x^2)+(Bx+C)(1-x)}\]
Expand the expressions
First, expand the term containing A:
\[A(1+x^2)=A+Ax^2\]
Next, expand the term containing B and C:
\[\begin{aligned}(Bx+C)(1-x)&=Bx+C-Bx^2-Cx\\ &=-Bx^2+(B-C)x+C\end{aligned}\]
Substituting these expressions into the numerator identity:
\[2=A+Ax^2-Bx^2+(B-C)x+C\]
Collect like terms
Grouping the coefficients of x², x and the constant terms:
\[2=(A-B)x^2+(B-C)x+(A+C)\]
Compare coefficients
Comparing the coefficients of x² on both sides:
\[A-B=0\]
Comparing the coefficients of x on both sides:
\[B-C=0\]
Comparing the constant terms on both sides:
\[A+C=2\]
Thus, we obtain three simultaneous equations:
\[ \begin{align} A-B&=0 \tag{1}\\ B-C&=0 \tag{2}\\ A+C&=2 \tag{3} \end{align} \]
Determine the values of A, B and C.
From equation (1):
\[A-B=0\]
Therefore:
\[A=B \tag{4}\]
From equation (2):
\[B-C=0\]
Therefore:
\[B=C \tag{5}\]
Combining equations (4) and (5):
\[A=B=C\]
Substituting A = C into equation (3):
\[A+C=2\]

\[A+A=2\]

\[2A=2\]
Therefore:
\[\boxed{A=1}\]
Since A = B = C:
\[\boxed{B=1,\qquad C=1}\]
Thus, the required constants are:
\[\boxed{A=1,\quad B=1,\quad C=1}\]
Write the partial fraction decomposition
Substituting the values of A, B and C:
\[\boxed{\frac{2}{(1-x)(1+x^2)}=\frac{1}{1-x}+\frac{x+1}{1+x^2}}\]
Substituting the decomposition into the integral, we have:
\[I=\int\frac{2}{(1-x)(1+x^2)}\,dx\]
Using the partial fraction decomposition:
\[I=\int\left[\frac{1}{1-x}+\frac{x+1}{1+x^2}\right]dx\]
Separating the numerator of the second fraction:
\[I=\int\left[\frac{1}{1-x}+\frac{x}{1+x^2}+\frac{1}{1+x^2}\right]dx\]
Applying the linearity property of integration:
\[ \begin{aligned} I={}& \int\frac{dx}{1-x}\\ &+\int\frac{x\,dx}{1+x^2}\\ &+\int\frac{dx}{1+x^2} \end{aligned} \]
Evaluate the first integral
Consider:
\[I_1=\int\frac{dx}{1-x}\]
Let:
\[t=1-x\]
Differentiating both sides:
\[dt=-dx\]
Therefore:
\[dx=-dt\]
Substituting into the integral:
\[I_1=-\int\frac{dt}{t}\]
Using the standard logarithmic integration formula:
\[I_1=-\log|t|+C\]
Substituting t = 1 - x:
\[\boxed{I_1=-\log|1-x|}\]
Important: The negative sign is essential because the derivative of 1 - x is -1.
Evaluate the second integral using substitution.
Consider:
\[I_2=\int\frac{x\,dx}{1+x^2}\]
Let:
\[t=1+x^2\]
Differentiating both sides:
\[dt=2x\,dx\]
Therefore:
\[x\,dx=\frac{dt}{2}\]
Substituting into the integral:
\[I_2=\frac12\int\frac{dt}{t}\]
Integrating:
\[I_2=\frac12\log|t|+C\]
Substituting t = 1 + x²:
\[I_2=\frac12\log|1+x^2|+C\]
Since 1 + x² is always positive for real x, the absolute value signs are unnecessary. Therefore:
\[\boxed{I_2=\frac12\log(1+x^2)}\]
Evaluate the third integral
Consider:
\[I_3=\int\frac{dx}{1+x^2}\]
Using the standard inverse trigonometric integration formula:
\[\boxed{\int\frac{dx}{1+x^2}=\tan^{-1}x+C}\]
Therefore:
\[\boxed{I_3=\tan^{-1}x}\]
Combine all the integrals
\[ \begin{aligned} I={}&-\log|1-x|\\ &+\frac12\log(1+x^2)\\ &+\tan^{-1}x+C \end{aligned} \]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int\frac{2}{(1-x)(1+x^2)}\,dx=-\log|1-x|+\frac12\log(1+x^2)+\tan^{-1}x+C}\]

🎯 Exam Significance

Exam Significance

›

The concepts applied in this question are relevant to JEE Main, JEE Advanced, BITSAT and other engineering entrance examinations.

Irreducible quadratic factors: Recognising when a quadratic expression cannot be factorised over the real numbers is essential for constructing the correct decomposition.
Linear numerator over a quadratic denominator: Students learn to use Bx + C in the numerator of the corresponding partial fraction.
Strategic decomposition: Breaking the numerator into x and constant terms simplifies the integration.
Substitution techniques: Recognising the derivative of 1 + x² helps identify logarithmic integrals quickly.
Inverse trigonometric functions: The integral of 1/(1 + x²) is fundamental in competitive mathematics.
Sign accuracy: Correctly handling the derivative of 1 - x prevents common errors in logarithmic integration.

🚫 Common Mistakes

Common Mistakes

›

Incorrect partial fraction structure: The numerator over the irreducible quadratic factor must be Bx + C.
Incorrect expansion: Remember that (Bx + C)(1 - x) = -Bx² + (B - C)x + C.
Incorrect coefficient comparison: The correct equations are A - B = 0, B - C = 0 and A + C = 2.
Missing negative sign: The integral of 1/(1 - x) is -log|1 - x|, not +log|1 - x|.
Incorrect substitution: If t = 1 + x², then dt = 2x dx, so x dx = dt/2.
Incorrect logarithmic integration: The integral of x/(1 + x²) is one-half log(1 + x²).
Confusing inverse tangent integration: The integral of 1/(1 + x²) is tan⁻¹x, not log(1 + x²).
Incorrect denominator transcription: Preserve the original factor (1 - x) throughout the calculation.
Forgetting the integration constant: Always include C in the final indefinite integral.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

A distinct linear factor and an irreducible quadratic factor require different numerator structures in partial fraction decomposition.
The numerator corresponding to an irreducible quadratic factor must be linear.
Coefficient comparison is a reliable method for determining unknown constants.
The integral of 1/(1 - x) carries a negative sign.
The integral of x/(1 + x²) is evaluated using logarithmic substitution.
The integral of 1/(1 + x²) produces an inverse tangent function.
Always verify the final antiderivative by differentiation.

📘 Concept & Theory

Theory and Concept

›

Integration of Rational Functions Containing Repeated Linear Factors

Partial fraction decomposition is a technique used to express a rational function as a sum of simpler fractions. These simpler fractions can then be integrated using standard integration formulae.

In the given problem, the denominator contains a repeated linear factor:

\[(x+2)^2\]

When the denominator contains a repeated linear factor of the form (x + a)², the appropriate partial fraction decomposition is:

\[\boxed{\frac{P(x)}{(x+a)^2}=\frac{A}{x+a}+\frac{B}{(x+a)^2}}\]

Each power of the repeated factor must be represented by a separate partial fraction.

Therefore, for the given problem, assume:

\[\boxed{\frac{3x-1}{(x+2)^2}=\frac{A}{x+2}+\frac{B}{(x+2)^2}}\]

Here, A and B are constants to be determined.

Important integration formulae:

Logarithmic integration:

\[\boxed{\int\frac{dx}{x+a}=\log|x+a|+C}\]

Integration of a repeated linear factor:

\[\boxed{\int\frac{dx}{(x+a)^2}=-\frac{1}{x+a}+C}\]

The first integral produces a logarithmic expression, whereas the second produces a rational expression.

🗺️ Solution Roadmap

Step-by-step Plan

›

Identify the repeated linear factor in the denominator.
Assume the appropriate partial fraction decomposition.
Multiply both sides by the common denominator.
Expand the expression and compare the coefficients of x and the constant term.
Determine the values of A and B.
Substitute the constants into the original integral.
Separate the expression into two simpler integrals.
Evaluate the first integral using logarithmic integration and the second using the power rule of integration.
Combine the results and verify the final answer by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 30 steps

Given —
\[I=\int\frac{3x-1}{(x+2)^2}\,dx\]
Apply partial fraction decomposition
Since the denominator contains the repeated linear factor (x + 2)², assume:
\[\frac{3x-1}{(x+2)^2}=\frac{A}{x+2}+\frac{B}{(x+2)^2}\]
Taking the LCM on the right-hand side:
\[\frac{3x-1}{(x+2)^2}=\frac{A(x+2)+B}{(x+2)^2}\]
Since the denominators on both sides are identical, equating the numerators gives:
\[\boxed{3x-1=A(x+2)+B}\]
Expand the expression
Expanding the right-hand side:
\[3x-1=Ax+2A+B\]
Compare coefficients
Comparing the coefficients of x on both sides:
\[A=3 \tag{1}\]
Comparing the constant terms:
\[2A+B=-1 \tag{2}\]
Thus, the two equations are:
\[ \begin{align}A&=3 \tag{1}\\ 2A+B&=-1 \tag{2}\end{align}\]
Determine the values of A and B.
From equation (1):
\[\boxed{A=3}\]
Substituting A = 3 into equation (2):
\[2(3)+B=-1\]
Simplifying:
\[6+B=-1\]
Transposing 6 to the right-hand side:
\[B=-1-6\]
Therefore:
\[\boxed{B=-7}\]
Thus, the required constants are:
\[\boxed{A=3,\qquad B=-7}\]
Write the partial fraction decomposition
Substituting the values of A and B:
\[\boxed{\frac{3x-1}{(x+2)^2}=\frac{3}{x+2}-\frac{7}{(x+2)^2}}\]
Substitute the decomposition into the integral
We have:
\[I=\int\frac{3x-1}{(x+2)^2}\,dx\]
Using the partial fraction decomposition:
\[I=\int\left[\frac{3}{x+2}-\frac{7}{(x+2)^2}\right]dx\]
Applying the linearity property of integration:
\[I=3\int\frac{dx}{x+2}-7\int\frac{dx}{(x+2)^2}\]
Evaluate the first integral
Consider:
\[I_1=3\int\frac{dx}{x+2}\]
Using the standard logarithmic integration formula:
\[\int\frac{dx}{x+a}=\log|x+a|+C\]
We obtain:
\[\begin{aligned}I_1&=3\log|x+2|\end{aligned}\]
Therefore:
\[\boxed{I_1=3\log|x+2|}\]
Evaluate the second integral using substitution
Consider:
\[I_2=-7\int\frac{dx}{(x+2)^2}\]
Let:
\[t=x+2\]
Differentiating both sides:
\[dt=dx\]
Substituting into the integral:
\[I_2=-7\int\frac{dt}{t^2}\]
Expressing the negative power:
\[I_2=-7\int t^{-2}\,dt\]
Using the power rule of integration:
\[\int t^n\,dt=\frac{t^{n+1}}{n+1}+C,\quad n\neq-1\]
Putting n = -2:
\[ \begin{aligned} I_2 &=-7\left(\frac{t^{-2+1}}{-2+1}\right)\\ &=-7\left(\frac{t^{-1}}{-1}\right)\\ &=7t^{-1}\\ &=\frac7t \end{aligned} \]
Substituting t = x + 2:
\[\boxed{I_2=\frac7{x+2}}\]
Combine both integrals
\[ \begin{aligned} I&=I_1+I_2+C\\ &=3\log|x+2|+\frac7{x+2}+C \end{aligned} \]
Therefore,
\[\boxed{\displaystyle\int\frac{3x-1}{(x+2)^2}\,dx=3\log|x+2|+\frac7{x+2}+C}\]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int\frac{3x-1}{(x+2)^2}\,dx=3\log|x+2|+\frac7{x+2}+C}\]

🎯 Exam Significance

Exam Significance

›

This problem is useful for CBSE Class 12 Mathematics board examination preparation because it demonstrates the integration of rational functions containing repeated linear factors.

Repeated linear factors: It tests the correct construction of partial fraction decomposition when a factor occurs more than once.
Coefficient comparison: Students practise determining unknown constants through algebraic identities.
Logarithmic integration: It reinforces the standard integral of the reciprocal of a linear expression.
Power rule: It develops understanding of integration involving negative powers.
Sign accuracy: It highlights the importance of handling negative coefficients correctly when integrating.
Stepwise presentation: It demonstrates a systematic method for solving rational integration problems in descriptive examinations.

Significance for Competitive Entrance Examinations

The techniques applied in this question are relevant to JEE Main, JEE Advanced, BITSAT and other engineering entrance examinations.

Repeated-factor decomposition: Understanding the structure of partial fractions involving repeated linear factors is essential for more complex rational integrals.
Algebraic simplification: Students learn to convert a complicated rational expression into simpler terms.
Negative power integration: The problem reinforces the power rule for negative exponents.
Logarithmic functions: It strengthens the application of logarithmic integration formulae.
Sign management: Accurate handling of negative coefficients is essential in objective-type questions.
Verification skills: Differentiation provides a reliable method to check an antiderivative.

🚫 Common Mistakes

Common Mistakes

›

Incorrect partial fraction structure: A repeated factor (x + 2)² requires two terms: A/(x + 2) and B/(x + 2)².
Incorrect coefficient comparison: The correct equations are A = 3 and 2A + B = -1.
Sign error in B: Remember that 6 + B = -1 gives B = -7.
Incorrect power rule: The integral of (x + 2)⁻² is -1/(x + 2), not +1/(x + 2).
Missing negative coefficient: The second integral has the coefficient -7. Multiplying it by the negative result of integration produces a positive rational term.
Incorrect logarithmic argument: The first integral produces log|x + 2|, not log|x + 1|.
Incorrect transcription: Preserve the square in the original denominator (x + 2)².
Forgetting the integration constant: Always include C in the final indefinite integral.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

A repeated linear factor requires a separate partial fraction for every power of that factor.
The correct decomposition for (x + 2)² contains two terms.
Coefficient comparison provides a systematic method for determining unknown constants.
The integral of 1/(x + 2) produces a logarithmic expression.
The integral of 1/(x + 2)² produces a negative rational expression.
The negative coefficient in the second partial fraction changes the resulting rational term to positive.
Always verify the final antiderivative by differentiation.

📘 Concept & Theory

Theory and Concept

›

Integration of Rational Functions Using Factorisation and Partial Fractions

A rational function is an expression in which both the numerator and denominator are polynomials. When the denominator can be factorised into simpler algebraic factors, partial fraction decomposition can be used to express the rational function as a sum of simpler fractions.

In this problem, the denominator is a difference of two squares:

\[x^4-1=(x^2)^2-1^2\]

Using the identity:

\[a^2-b^2=(a-b)(a+b)\]

we obtain:

\[x^4-1=(x^2-1)(x^2+1)\]

Further factorisation gives:

\[\boxed{x^4-1=(x-1)(x+1)(x^2+1)}\]

The denominator therefore contains two distinct linear factors and one irreducible quadratic factor.

The appropriate partial fraction decomposition is:

\[\frac{1}{(x-1)(x+1)(x^2+1)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{Cx+D}{x^2+1}\]

Alternatively, we can first use the substitution \(y=x^2\) to simplify the algebraic expression and then decompose the resulting quadratic factors. We will use this approach in the solution.

🗺️ Solution Roadmap

Step-by-step Plan

›

Substitute \(y=x^2\) to simplify the denominator algebraically.
Factorise \(y^2-1\) using the difference of squares identity.
Apply partial fraction decomposition in terms of y.
Determine the constants by comparing coefficients.
Substitute \(y=x^2\) back into the decomposition.
Factorise \(x^2-1\) further into linear factors.
Integrate the resulting logarithmic and inverse-trigonometric terms separately.
Combine the results and verify the answer by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 44 steps

Given —
\[I=\int\frac{1}{x^4-1}\,dx\]
Apply substitution
Let:
\[y=x^2\]
Then:
\[x^4=y^2\]
Therefore, the rational expression becomes:
\[\frac{1}{x^4-1}=\frac{1}{y^2-1}\]
Using the difference of squares identity:
\[y^2-1=(y+1)(y-1)\]
Hence:
\[\boxed{\frac{1}{y^2-1}=\frac{1}{(y+1)(y-1)}}\]
Apply partial fraction decomposition
Assume:
\[\frac{1}{(y+1)(y-1)}=\frac{A}{y+1}+\frac{B}{y-1}\]
Taking the LCM on the right-hand side:
\[\frac{1}{(y+1)(y-1)}=\frac{A(y-1)+B(y+1)}{(y+1)(y-1)}\]
Equating the numerators:
\[1=A(y-1)+B(y+1)\]
Expand the expression
Expanding the right-hand side:
\[1=Ay-A+By+B\]
Rearranging:
\[1=(A+B)y+(-A+B)\]
Compare coefficients
Comparing the coefficients of y on both sides:
\[A+B=0 \tag{1}\]
Comparing the constant terms:
\[-A+B=1 \tag{2}\]
From equation (1):
\[A=-B\]
Substituting this value into equation (2):
\[-(-B)+B=1\]
Simplifying:
\[2B=1\]
Therefore:
\[\boxed{B=\frac12}\]
Since \(A=-B\):
\[\boxed{A=-\frac12}\]
Thus, the required partial fraction decomposition is:
\[\boxed{\frac{1}{y^2-1}=\frac{1}{2(y-1)}-\frac{1}{2(y+1)}}\]
Substitute \(y=x^2\) back into the expression
Replacing y with \(x^2\):
\[\frac{1}{x^4-1}=\frac{1}{2(x^2-1)}-\frac{1}{2(x^2+1)}\]
Therefore:
\[I=\frac12\int\frac{dx}{x^2-1}-\frac12\int\frac{dx}{x^2+1}\]
Evaluate the first integral
Consider:
\[I_1=\frac12\int\frac{dx}{x^2-1}\]
Factorising the denominator:
\[x^2-1=(x-1)(x+1)\]
Thus:
\[I_1=\frac12\int\frac{dx}{(x-1)(x+1)}\]
Applying partial fraction decomposition, assume:
\[\frac{1}{(x-1)(x+1)}=\frac{P}{x-1}+\frac{Q}{x+1}\]
Taking the LCM:
\[1=P(x+1)+Q(x-1)\]
Expanding:
\[1=Px+P+Qx-Q\]
Rearranging:
\[1=(P+Q)x+(P-Q)\]
Comparing coefficients:
\[P+Q=0\]

\[P-Q=1\]
From the first equation:
\[Q=-P\]
Substituting into the second equation:
\[P-(-P)=1\]
Therefore:
\[2P=1\]
Hence:
\[P=\frac12,\qquad Q=-\frac12\]
Consequently:
\[\frac{1}{x^2-1}=\frac12\left(\frac{1}{x-1}-\frac{1}{x+1}\right)\]
Substituting into \(I_1\):
\[I_1=\frac14\int\left(\frac{1}{x-1}-\frac{1}{x+1}\right)dx\]
Separating the integrals:
\[I_1=\frac14\int\frac{dx}{x-1}-\frac14\int\frac{dx}{x+1}\]
Using the standard logarithmic integration formula:
\[\int\frac{dx}{x-a}=\log|x-a|+C\]
We obtain:
\[\begin{aligned}I_1&=\frac14\log|x-1|-\frac14\log|x+1|\\ &=\frac14\log\left|\frac{x-1}{x+1}\right|\end{aligned}\]
Therefore:
\[\boxed{I_1=\frac14\log\left|\frac{x-1}{x+1}\right|}\]
Evaluate the second integral
Consider:
\[I_2=-\frac12\int\frac{dx}{x^2+1}\]
Using the standard inverse-trigonometric integration formula:
\[\boxed{\int\frac{dx}{1+x^2}=\tan^{-1}x+C}\]
We obtain:
\[I_2=-\frac12\tan^{-1}x\]
Therefore:
\[\boxed{I_2=-\frac12\tan^{-1}x}\]
Combine both integrals
We have:
\[I=I_1+I_2+C\]
Substituting the values of \(I_1\) and \(I_2\):
\[I=\frac14\log\left|\frac{x-1}{x+1}\right|-\frac12\tan^{-1}x+C\]
Hence, the required integral is:
\[\boxed{\displaystyle\int\frac{dx}{x^4-1}=\frac14\log\left|\frac{x-1}{x+1}\right|-\frac12\tan^{-1}x+C}\]

💡 Answer

Final Answer

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\[\boxed{\displaystyle\int\frac{dx}{x^4-1}=\frac14\log\left|\frac{x-1}{x+1}\right|-\frac12\tan^{-1}x+C}\]

🎯 Exam Significance

Exam Significance

›

This problem is important for CBSE Class 12 Mathematics because it combines algebraic factorisation, partial fraction decomposition and standard integration formulae.

Factorisation: It tests the application of the difference of squares identity to higher powers.
Partial fractions: It reinforces the decomposition of rational functions into simpler fractions.
Logarithmic integration: It develops the ability to integrate fractions with linear denominators.
Inverse-trigonometric integration: It tests recognition of the standard integral involving \(1/(1+x^2)\).
Algebraic accuracy: It demonstrates why a quadratic denominator must sometimes be factorised further before integration.
Stepwise presentation: The solution provides a clear sequence of algebraic and integration steps suitable for descriptive examination answers.

Significance for Competitive Entrance Examinations

The concepts used in this problem are relevant to JEE Main, JEE Advanced, BITSAT and other competitive entrance examinations.

Algebraic transformation: Substitution can simplify the structure of a rational expression.
Partial fraction decomposition: The method is frequently used to evaluate rational integrals.
Mixed integration techniques: The problem combines logarithmic and inverse-trigonometric integration in a single expression.
Factorisation skills: Recognising the complete factorisation of a polynomial helps avoid unnecessary calculations.
Conceptual precision: It tests whether students understand when a logarithmic formula can be applied directly and when further decomposition is necessary.
Verification: Differentiation can be used to check an answer and identify errors in coefficients or signs.

🚫 Common Mistakes

Common Mistakes

›

Incorrect signs in partial fractions: For the decomposition in y, the correct constants are \(A=-\frac12\) and \(B=\frac12\).
Incorrect logarithmic integration: The integral of \(1/(x^2-1)\) is not simply \(\log|x^2-1|\).
Missing factor of 2: Further decomposition of \(1/(x^2-1)\) introduces an additional factor of \(1/2\), giving an overall logarithmic coefficient of \(1/4\).
Confusing quadratic forms: The integral of \(1/(x^2+1)\) gives \(\tan^{-1}x\), whereas \(1/(x^2-1)\) requires partial fractions.
Ignoring absolute values: Logarithmic terms involving real linear factors should be written with absolute values.
Incorrect use of substitution: Substituting \(y=x^2\) simplifies the algebraic decomposition, but it does not mean that \(dy=dx\). The original integral is still with respect to x.
Forgetting the integration constant: Always include C in the final indefinite integral.

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 7 points

The denominator \(x^4-1\) can be factorised as \((x^2-1)(x^2+1)\).
The substitution \(y=x^2\) simplifies the initial partial fraction decomposition.
The correct decomposition is \(1/(y^2-1)=1/[2(y-1)]-1/[2(y+1)]\).
The factor \(x^2-1\) must be decomposed further into the linear factors \(x-1\) and \(x+1\).
The logarithmic contribution is \(\frac14\log|(x-1)/(x+1)|\).
The irreducible quadratic factor contributes the inverse-tangent term \(-\frac12\tan^{-1}x\).
Differentiation is an effective way to verify the final antiderivative.

📘 Concept & Theory

Theory and Concept

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Integration Using Substitution and Partial Fractions

This problem involves a rational function containing the product of x and a polynomial expression of the form \(x^n+1\).

The expression can be simplified by substituting:

\[ t=x^n \]

Differentiating both sides:

\[ dt=nx^{n-1}\,dx \]

Dividing by \(nt\), and using \(t=x^n\):

\[ \frac{dt}{nt}=\frac{dx}{x} \]

This substitution converts the original integral into a simpler rational integral involving \(t\), which can be evaluated using partial fraction decomposition.

Important integration formula:

\[\boxed{\int\frac{dt}{t(t+1)}=\log|t|-\log|t+1|+C}\]

🗺️ Solution Roadmap

Step-by-step Plan

›

Identify the expression \(x^n\) and substitute \(t=x^n\).
Differentiate the substitution to obtain a relationship between \(dx/x\) and \(dt/t\).
Transform the original integral into an integral in t.
Apply partial fraction decomposition.
Determine the unknown constants by comparing coefficients.
Integrate the resulting simpler fractions.
Substitute \(t=x^n\) back into the result.
Verify the final answer by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 31 steps

Given —
\[I=\int\frac{1}{x(x^n+1)}\,dx\]
Apply substitution
Let:
\[t=x^n\]
Differentiating both sides with respect to x:
\[\frac{dt}{dx}=nx^{n-1}\]
Therefore:
\[dt=nx^{n-1}\,dx\]
Multiplying and dividing by x:
\[dt=nx^n\frac{dx}{x}\]
Since \(x^n=t\):
\[dt=nt\frac{dx}{x}\]
Rearranging:
\[\boxed{\frac{dx}{x}=\frac{dt}{nt}}\]
Transform the original integral
We have:
\[I=\int\frac{1}{x(x^n+1)}\,dx\]
Rearranging the integrand:
\[I=\int\frac{1}{x^n+1}\cdot\frac{dx}{x}\]
Substituting \(x^n=t\) and \(\frac{dx}{x}=\frac{dt}{nt}\):
\[I=\int\frac{1}{t+1}\cdot\frac{dt}{nt}\]
Taking the constant factor outside:
\[\boxed{I=\frac1n\int\frac{dt}{t(t+1)}}\]
Apply partial fraction decomposition
Assume:
\[\frac{1}{t(t+1)}=\frac{A}{t}+\frac{B}{t+1}\]
Taking the LCM on the right-hand side:
\[\frac{1}{t(t+1)}=\frac{A(t+1)+Bt}{t(t+1)}\]
Equating the numerators:
\[1=A(t+1)+Bt\]
Expanding:
\[1=At+A+Bt\]
Rearranging:
\[1=(A+B)t+A\]
Determine the values of A and B
Comparing the coefficients of t:
\[A+B=0 \tag{1}\]
Comparing the constant terms:
\[A=1 \tag{2}\]
From equation (2):
\[A=1\]
Substituting A = 1 into equation (1):
\[1+B=0\]
Therefore:
\[B=-1\]
Hence:
\[\boxed{A=1,\qquad B=-1}\]
The partial fraction decomposition is:
\[\boxed{\frac{1}{t(t+1)}=\frac1t-\frac1{t+1}}\]
Integrate the partial fractions
Substituting the decomposition into the integral:
\[I=\frac1n\int\left(\frac1t-\frac1{t+1}\right)dt\]
Separating the integrals:
\[I=\frac1n\left[\int\frac{dt}{t}-\int\frac{dt}{t+1}\right]\]
Using the standard logarithmic integration formula:
\[\int\frac{dt}{t}=\log|t|+C\]
Similarly:
\[\int\frac{dt}{t+1}=\log|t+1|+C\]
Therefore:
\[I=\frac1n\left[\log|t|-\log|t+1|\right]+C\]
Combining the logarithms:
\[I=\frac1n\log\left|\frac{t}{t+1}\right|+C\]
Substitute the original variable
Since:
\[t=x^n\]
Substituting back:
\[I=\frac1n\log\left|\frac{x^n}{x^n+1}\right|+C\]
Hence, the required integral is:
\[\boxed{\displaystyle\int\frac{1}{x(x^n+1)}\,dx=\frac1n\log\left|\frac{x^n}{x^n+1}\right|+C}\]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int\frac{1}{x(x^n+1)}\,dx=\frac1n\log\left|\frac{x^n}{x^n+1}\right|+C}\]

🎯 Exam Significance

Exam Significance

›

Substitution: Tests the ability to identify a suitable substitution involving a power of x.
Partial fractions: Reinforces the decomposition of a rational function into simpler fractions.
Logarithmic integration: Tests the application of standard logarithmic integration formulae.
Generalisation: Introduces a result involving the parameter n rather than a fixed numerical exponent.
Algebraic accuracy: Requires careful handling of the factor \(1/n\) introduced during substitution.

Significance for Competitive Entrance Examinations

Parameter-based integration: Develops the ability to solve integrals involving a general positive integer n.
Substitution techniques: Strengthens the recognition of expressions suitable for logarithmic differentiation.
Partial fraction decomposition: Provides a foundation for more complicated rational integrals.
Algebraic simplification: Demonstrates how a complicated-looking integrand can be reduced to a standard form.
Derivative-based verification: Encourages checking antiderivatives through differentiation.

🚫 Common Mistakes

Common Mistakes

›

Forgetting the factor n while differentiating \(t=x^n\).
Writing \(dx/x=dt/n\) instead of the correct expression \(dx/x=dt/(nt)\).
Using incorrect signs in the partial fraction decomposition.
Forgetting the factor \(1/n\) outside the transformed integral.
Combining logarithms without retaining the absolute value.
Substituting \(t=x^n\) back incorrectly.
Omitting the integration constant C.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

The substitution \(t=x^n\) simplifies the given rational integral.
The differential relationship is \(dx/x=dt/(nt)\).
The transformed rational function can be decomposed as \(1/[t(t+1)]=1/t-1/(t+1)\).
The resulting integral is evaluated using logarithmic integration.
The factor \(1/n\) must be retained throughout the calculation.
Differentiation confirms the correctness of the final result.

📘 Concept & Theory

Theory and Concept

›

Integration of Rational Functions Using Polynomial Division and Partial Fractions

A rational function is a quotient of two polynomials. When the degree of the numerator is equal to the degree of the denominator, polynomial division can be used to express the rational function as the sum of a constant and a proper rational function.

In this problem, the numerator and denominator are both polynomials of degree four:

\[(x^2+1)(x^2+2)\]

and

\[(x^2+3)(x^2+4)\]

Therefore, we first divide the numerator by the denominator. The resulting proper rational function can then be expressed using partial fractions.

We will use the following standard integration formula:

\[\boxed{\int\frac{dx}{x^2+a^2}=\frac1a\tan^{-1}\left(\frac xa\right)+C}\]

This formula is applicable because the remaining denominators contain expressions of the form \(x^2+a^2\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Introduce \(y=x^2\) to simplify the algebraic expressions.
Expand the numerator and denominator.
Perform polynomial division to separate the constant term from the proper rational function.
Factorise the denominator into linear factors in y.
Apply partial fraction decomposition.
Determine the unknown constants by comparing coefficients.
Substitute \(y=x^2\) back into the expression.
Integrate the constant term and the two quadratic-denominator terms separately.
Apply the inverse-tangent integration formula and combine the results.
Verify the final answer by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 46 steps

Given —
\[I=\int\frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}\,dx\]
Introduce a substitution for algebraic simplification.
Let:
\[y=x^2\]
Then the rational expression becomes:
\[\frac{(y+1)(y+2)}{(y+3)(y+4)}\]
Expanding the numerator:
\[\begin{aligned}(y+1)(y+2)&=y^2+2y+y+2\\ &=y^2+3y+2\end{aligned}\]
Expanding the denominator:
\[\begin{aligned}(y+3)(y+4)&=y^2+4y+3y+12\\ &=y^2+7y+12\end{aligned}\]
Therefore:
\[\boxed{\frac{(y+1)(y+2)}{(y+3)(y+4)}=\frac{y^2+3y+2}{y^2+7y+12}}\]
Apply polynomial division
Since the numerator and denominator have the same degree, divide the numerator by the denominator.
Observe that:
\[y^2+7y+12-(y^2+3y+2)\]
Simplifying:
\[\begin{aligned}&=y^2+7y+12-y^2-3y-2\\ &=4y+10\end{aligned}\]
Thus:
\[\begin{aligned}\frac{y^2+3y+2}{y^2+7y+12}&=1-\frac{4y+10}{y^2+7y+12}\\ &=1-\frac{4y+10}{(y+3)(y+4)}\end{aligned}\]
Hence:
\[\boxed{\frac{(y+1)(y+2)}{(y+3)(y+4)}=1-\frac{4y+10}{(y+3)(y+4)}}\]
Apply partial fraction decomposition.
Consider the proper rational function:
\[\frac{4y+10}{(y+3)(y+4)}\]
Assume:
\[\frac{4y+10}{(y+3)(y+4)}=\frac{A}{y+3}+\frac{B}{y+4}\]
Taking the LCM on the right-hand side:
\[\frac{4y+10}{(y+3)(y+4)}=\frac{A(y+4)+B(y+3)}{(y+3)(y+4)}\]
Equating the numerators:
\[4y+10=A(y+4)+B(y+3)\]
Expanding the right-hand side:
\[\begin{aligned}4y+10&=Ay+4A+By+3B\\ &=(A+B)y+(4A+3B)\end{aligned}\]
Compare coefficients
Comparing the coefficients of y:
\[A+B=4 \tag{1}\]
Comparing the constant terms:
\[4A+3B=10 \tag{2}\]
From equation (1):
\[A=4-B\]
Substituting this value into equation (2):
\[4(4-B)+3B=10\]
Expanding:
\[16-4B+3B=10\]
Simplifying:
\[16-B=10\]
Therefore:
\[-B=10-16\]
Hence:
\[\boxed{B=6}\]
Since \(A=4-B\):
\[A=4-6\]
Therefore:
\[\boxed{A=-2}\]
Thus, the required constants are:
\[\boxed{A=-2,\qquad B=6}\]
Write the partial fraction decomposition
Substituting the values of A and B:
\[\boxed{\frac{4y+10}{(y+3)(y+4)}=-\frac2{y+3}+\frac6{y+4}}\]
Recall that:
\[\frac{(y+1)(y+2)}{(y+3)(y+4)}=1-\frac{4y+10}{(y+3)(y+4)}\]
Therefore:
\[\begin{aligned}\frac{(y+1)(y+2)}{(y+3)(y+4)}&=1-\left(-\frac2{y+3}+\frac6{y+4}\right)\\&=1+\frac2{y+3}-\frac6{y+4}\end{aligned}\]
Hence:
\[\boxed{\frac{(y+1)(y+2)}{(y+3)(y+4)}=1+\frac2{y+3}-\frac6{y+4}}\]
Substitute \(y=x^2\) back into the expression
Replacing y with \(x^2\):
\[\boxed{\frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}=1+\frac2{x^2+3}-\frac6{x^2+4}}\]
Therefore, the original integral becomes:
\[I=\int\left(1+\frac2{x^2+3}-\frac6{x^2+4}\right)dx\]
Applying the linearity property of integration:
\[I=\int dx+2\int\frac{dx}{x^2+3}-6\int\frac{dx}{x^2+4}\]
Evaluate the first integral
We have:
\[I_1=\int dx\]
Using the standard formula:
\[\boxed{I_1=x}\]
Evaluate the second integral
Consider:
\[I_2=2\int\frac{dx}{x^2+3}\]
Writing the denominator in the form \(x^2+a^2\):
\[x^2+3=x^2+(\sqrt3)^2\]
Using the standard formula:
\[\int\frac{dx}{x^2+a^2}=\frac1a\tan^{-1}\left(\frac xa\right)+C\]
Here, \(a=\sqrt3\). Therefore:
\[\begin{aligned}I_2&=2\left[\frac1{\sqrt3}\tan^{-1}\left(\frac{x}{\sqrt3}\right)\right]\\ &=\frac2{\sqrt3}\tan^{-1}\left(\frac{x}{\sqrt3}\right)\end{aligned}\]
Hence:
\[\boxed{I_2=\frac2{\sqrt3}\tan^{-1}\left(\frac{x}{\sqrt3}\right)}\]
Evaluate the third integral
Consider:
\[I_3=-6\int\frac{dx}{x^2+4}\]
Writing the denominator in the form \(x^2+a^2\):
\[x^2+4=x^2+2^2\]
Here, \(a=2\). Applying the standard integration formula:
\[\begin{aligned}I_3&=-6\left[\frac12\tan^{-1}\left(\frac x2\right)\right]\\ &=-3\tan^{-1}\left(\frac x2\right)\end{aligned}\]
Hence:
\[\boxed{I_3=-3\tan^{-1}\left(\frac x2\right)}\]
Combine all three integrals
We have:
\[I=I_1+I_2+I_3+C\]
Substituting the results obtained above:
\[\begin{aligned}I&=x+\frac2{\sqrt3}\tan^{-1}\left(\frac{x}{\sqrt3}\right)-3\tan^{-1}\left(\frac x2\right)+C\end{aligned}\]
Therefore, the required integral is:
\[\boxed{\displaystyle\int\frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}\,dx=x+\frac2{\sqrt3}\tan^{-1}\left(\frac{x}{\sqrt3}\right)-3\tan^{-1}\left(\frac x2\right)+C}\]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int\frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}\,dx=x+\frac2{\sqrt3}\tan^{-1}\left(\frac{x}{\sqrt3}\right)-3\tan^{-1}\left(\frac x2\right)+C}\]

🎯 Exam Significance

Exam Significance

›

This problem is useful for CBSE Class 12 Mathematics board examination preparation because it combines polynomial division, partial fraction decomposition and inverse-trigonometric integration.

Polynomial division: It tests the ability to simplify a rational function when the numerator and denominator have equal degrees.
Partial fractions: It develops the ability to decompose a proper rational function into simpler fractions.
Inverse-trigonometric integration: It reinforces the standard formula involving \(1/(x^2+a^2)\).
Algebraic accuracy: It tests the correct expansion and subtraction of polynomial expressions.
Coefficient comparison: It provides practice in finding unknown constants in partial fraction decomposition.
Stepwise presentation: It demonstrates how to organise a multi-stage integration problem clearly in a descriptive examination.

Significance for Competitive Entrance Examinations

The techniques applied in this problem are relevant to JEE Main, JEE Advanced, BITSAT and other competitive entrance examinations.

Improper rational functions: Students learn to recognise when polynomial division must precede partial fraction decomposition.
Algebraic manipulation: The problem develops accuracy in simplifying rational expressions.
Partial fraction techniques: It strengthens the ability to break complicated expressions into standard integrable forms.
Inverse-tangent functions: It tests recognition and application of inverse-trigonometric integration formulae.
Combined techniques: It demonstrates how polynomial division, partial fractions and standard integration formulae work together.
Verification skills: Differentiation provides a reliable method for checking an antiderivative.

🚫 Common Mistakes

Common Mistakes

›

Incorrect polynomial subtraction: The difference between \(y^2+7y+12\) and \(y^2+3y+2\) is \(4y+10\), not \(y+10\).
Incorrect partial fraction constants: For \((4y+10)/[(y+3)(y+4)]\), the correct constants are \(A=-2\) and \(B=6\).
Incorrect signs after subtraction: Since the decomposition is subtracted from 1, the final expression becomes \(1+2/(y+3)-6/(y+4)\).
Incorrect inverse-tangent coefficient: The formula requires the factor \(1/a\). Thus, the coefficient of the second inverse-tangent term is \(-3\), not \(+3\).
Incorrect handling of substitution: The substitution \(y=x^2\) is used only to simplify the algebraic expression. The integration variable remains x.
Forgetting the integration constant: Always include C in the final indefinite integral.
Skipping verification: Differentiating the final expression helps identify errors in partial fraction constants and signs.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

When the numerator and denominator have equal degrees, polynomial division should be considered first.
The difference between the denominator and numerator in this problem is \(4y+10\).
The correct partial fraction decomposition is:
\[ \frac{4y+10}{(y+3)(y+4)} = -\frac2{y+3}+\frac6{y+4} \]
Subtracting this decomposition from 1 gives:
\[ 1+\frac2{y+3}-\frac6{y+4} \]
The integral of \(1/(x^2+a^2)\) produces an inverse-tangent function.
The coefficients of the inverse-tangent terms must be calculated carefully.
Differentiation confirms the correctness of the final result.

📘 Concept & Theory

Theory and Concept

›

Integration Using Substitution and Partial Fraction Decomposition

This problem involves a rational function in which the numerator is \(2x\), while the denominator contains quadratic expressions in \(x\).

Notice that the derivative of \(x^2\) is \(2x\). Therefore, the substitution:

\[t=x^2\]

is appropriate because it converts the numerator and differential into a single differential term.

Differentiating:

\[dt=2x\,dx\]

The integral consequently becomes a rational function of t:

\[\int\frac{dt}{(t+1)(t+3)}\]

This expression can be evaluated conveniently using partial fraction decomposition.

Important formula:

\[\boxed{\int\frac{dt}{(t+a)(t+b)}=\frac{1}{b-a}\log\left|\frac{t+a}{t+b}\right|+C}\]

Here, \(a\) and \(b\) are distinct constants.

🗺️ Solution Roadmap

Step-by-step Plan

›

Observe that the numerator \(2x\) is the derivative of \(x^2\).
Substitute \(t=x^2\) and obtain \(dt=2x\,dx\).
Rewrite the denominator in terms of t.
Apply partial fraction decomposition.
Determine the unknown constants by comparing coefficients.
Integrate the resulting simple fractions.
Substitute \(t=x^2\) back into the expression.
Verify the final answer by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 30 steps

Given —
\[I=\int\frac{2x}{(x^2+1)(x^2+3)}\,dx\]
Apply substitution
Let:
\[t=x^2\]
Differentiating both sides with respect to x:
\[dt=2x\,dx\]
Also, since \(t=x^2\):
\[x^2+1=t+1\]
and
\[x^2+3=t+3\]
Transform the original integral
Substituting the above expressions:
\[\begin{aligned}I&=\int\frac{2x\,dx}{(x^2+1)(x^2+3)}\\ &=\int\frac{dt}{(t+1)(t+3)}\end{aligned}\]
Therefore:
\[\boxed{I=\int\frac{dt}{(t+1)(t+3)}}\]
Apply partial fraction decomposition
Assume:
\[\frac{1}{(t+1)(t+3)}=\frac{A}{t+1}+\frac{B}{t+3}\]
Taking the LCM on the right-hand side:
\[\frac{1}{(t+1)(t+3)}=\frac{A(t+3)+B(t+1)}{(t+1)(t+3)}\]
Equating the numerators:
\[1=A(t+3)+B(t+1)\]
Expand the expression
Expanding:
\[\begin{aligned}1&=At+3A+Bt+B\\ &=(A+B)t+(3A+B)\end{aligned}\]
Compare coefficients
Comparing the coefficients of t:
\[A+B=0 \tag{1}\]
Comparing the constant terms:
\[3A+B=1 \tag{2}\]
From equation (1):
\[B=-A\]
Substituting into equation (2):
\[3A-A=1\]
Simplifying:
\[2A=1\]
Therefore:
\[\boxed{A=\frac12}\]
Since \(B=-A\):
\[\boxed{B=-\frac12}\]
Hence, the required constants are:
\[\boxed{A=\frac12,\quad B=-\frac12}\]
Write the partial fraction decomposition
Substituting the values of A and B:
\[\boxed{\frac{1}{(t+1)(t+3)}=\frac{1}{2(t+1)}-\frac{1}{2(t+3)}}\]
Therefore:
\[I=\frac12\int\frac{dt}{t+1}-\frac12\int\frac{dt}{t+3}\]
Evaluate the first integral
Consider:
\[I_1=\frac12\int\frac{dt}{t+1}\]
Using the standard logarithmic integration formula:
\[\int\frac{dt}{t+a}=\log|t+a|+C\]
We obtain:
\[\boxed{I_1=\frac12\log|t+1|}\]
Evaluate the second integral
Consider:
\[I_2=-\frac12\int\frac{dt}{t+3}\]
Applying the same logarithmic integration formula:
\[\boxed{I_2=-\frac12\log|t+3|}\]
Combine both integrals
We have:
\[I=I_1+I_2+C\]
Substituting the results:
\[I=\frac12\log|t+1|-\frac12\log|t+3|+C\]
Taking the common factor:
\[I=\frac12\left[\log|t+1|-\log|t+3|\right]+C\]
Using the logarithmic identity:
\[\log a-\log b=\log\left|\frac ab\right|\]
Substitute the original variable
Since:
\[t=x^2\]
Substituting back:
\[I=\frac12\log\left|\frac{x^2+1}{x^2+3}\right|+C\]
Therefore, the required integral is:
\[\boxed{\displaystyle\int\frac{2x}{(x^2+1)(x^2+3)}\,dx=\frac12\log\left(\frac{x^2+1}{x^2+3}\right)+C}\]
Note that both \(x^2+1\) and \(x^2+3\) are positive for every real value of x, so absolute value signs are not necessary in this final logarithmic expression.

🎯 Exam Significance

Exam Significance

›

This problem is useful for CBSE Class 12 Mathematics board examination preparation because it demonstrates how substitution can simplify a rational integral before applying partial fraction decomposition.

Substitution: It tests the identification of an expression whose derivative appears in the numerator.
Partial fractions: It reinforces the decomposition of rational functions into simpler fractions.
Logarithmic integration: It tests the application of standard logarithmic integration formulae.
Algebraic accuracy: It develops the ability to compare coefficients and determine unknown constants.
Method selection: It demonstrates why choosing an appropriate substitution can make integration significantly simpler.

Significance for Competitive Entrance Examinations

The techniques applied in this problem are relevant to JEE Main, JEE Advanced, BITSAT and other competitive entrance examinations.

Recognition of derivatives: Students learn to identify \(2x\) as the derivative of \(x^2\).
Substitution techniques: It develops the ability to transform complicated rational integrals into simpler forms.
Partial fraction decomposition: It strengthens algebraic manipulation skills used in more advanced integration problems.
Logarithmic simplification: It reinforces the use of logarithmic identities.
Efficient problem-solving: It demonstrates how substitution can reduce lengthy calculations.

🚫 Common Mistakes

Common Mistakes

›

Incorrect partial fraction structure: Writing constant numerators directly over the original quadratic factors does not match the numerator \(2x\). Substitution is a more convenient approach here.
Incorrect differential: From \(t=x^2\), the correct differential is \(dt=2x\,dx\).
Missing substitution: Both quadratic expressions in the denominator must be converted into expressions involving t.
Incorrect partial fraction signs: The correct constants are \(A=\frac12\) and \(B=-\frac12\).
Incorrect logarithmic coefficients: Both logarithmic terms retain the factor \(\frac12\).
Incorrect substitution back: Replace t with \(x^2\) in both logarithmic terms.
Forgetting the integration constant: Always include C in the final indefinite integral.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

The numerator \(2x\) suggests the substitution \(t=x^2\).
The differential relationship is \(dt=2x\,dx\).
The transformed integral is \(\int dt/[(t+1)(t+3)]\).
The correct partial fraction decomposition is:
\[ \frac{1}{(t+1)(t+3)} = \frac{1}{2(t+1)} - \frac{1}{2(t+3)} \]
Both resulting terms are evaluated using logarithmic integration.
Recognising the derivative of an expression in the numerator can simplify a rational integral considerably.
Differentiation confirms the correctness of the final antiderivative.

📘 Concept & Theory

Theory and Concept

›

Integration Using Substitution and Partial Fraction Decomposition

This problem involves a rational function in which the numerator is \(2x\), while the denominator contains quadratic expressions in \(x\).

Notice that the derivative of \(x^2\) is \(2x\). Therefore, the substitution:

\[ t=x^2 \]

is appropriate because it converts the numerator and differential into a single differential term.

Differentiating:

\[ dt=2x\,dx \]

The integral consequently becomes a rational function of t:

\[ \int\frac{dt}{(t+1)(t+3)} \]

This expression can be evaluated conveniently using partial fraction decomposition.

Important formula:

\[\boxed{\int\frac{dt}{(t+a)(t+b)}=\frac{1}{b-a}\log\left|\frac{t+a}{t+b}\right|+C}\]

Here, \(a\) and \(b\) are distinct constants.

🗺️ Solution Roadmap

Step-by-step Plan

›

Observe that the numerator \(2x\) is the derivative of \(x^2\).
Substitute \(t=x^2\) and obtain \(dt=2x\,dx\).
Rewrite the denominator in terms of t.
Apply partial fraction decomposition.
Determine the unknown constants by comparing coefficients.
Integrate the resulting simple fractions.
Substitute \(t=x^2\) back into the expression.
Verify the final answer by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 32 steps

Given —
\[I=\int\frac{2x}{(x^2+1)(x^2+3)}\,dx\]
Apply substitution
Let:
\[t=x^2\]
Differentiating both sides with respect to x:
\[dt=2x\,dx\]
Also, since \(t=x^2\):
\[x^2+1=t+1\]
and
\[x^2+3=t+3\]
Transform the original integral
Substituting the above expressions:
\[ \begin{aligned} I &=\int\frac{2x\,dx}{(x^2+1)(x^2+3)}\\ &=\int\frac{dt}{(t+1)(t+3)} \end{aligned} \]
Therefore:
\[\boxed{I=\int\frac{dt}{(t+1)(t+3)}}\]
Apply partial fraction decomposition
Assume:
\[\frac{1}{(t+1)(t+3)}=\frac{A}{t+1}+\frac{B}{t+3}\]
Taking the LCM on the right-hand side:
\[\frac{1}{(t+1)(t+3)}=\frac{A(t+3)+B(t+1)}{(t+1)(t+3)}\]
Equating the numerators:
\[1=A(t+3)+B(t+1)\]
Expand the expression
Expanding:
\[\begin{aligned}1&=At+3A+Bt+B\\&=(A+B)t+(3A+B)\end{aligned}\]
Compare coefficients
Comparing the coefficients of t:
\[A+B=0 \tag{1}\]
Comparing the constant terms:
\[3A+B=1 \tag{2}\]
From equation (1):
\[B=-A\]
Substituting into equation (2):
\[3A-A=1\]
Simplifying:
\[2A=1\]
Therefore:
\[\boxed{A=\frac12}\]
Since \(B=-A\):
\[\boxed{B=-\frac12}\]
Hence, the required constants are:
\[ \boxed{ A=\frac12,\quad B=-\frac12 } \]
Write the partial fraction decomposition
Substituting the values of A and B:
\[\boxed{\frac{1}{(t+1)(t+3)}=\frac{1}{2(t+1)}-\frac{1}{2(t+3)}}\]
Therefore:
\[I=\frac12\int\frac{dt}{t+1}-\frac12\int\frac{dt}{t+3}\]
Evaluate the first integral
Consider:
\[I_1=\frac12\int\frac{dt}{t+1}\]
Using the standard logarithmic integration formula:
\[\int\frac{dt}{t+a}=\log|t+a|+C\]
We obtain:
\[\boxed{I_1=\frac12\log|t+1|}\]
Evaluate the second integral
Consider:
\[I_2=-\frac12\int\frac{dt}{t+3}\]
Applying the same logarithmic integration formula:
\[\boxed{I_2=-\frac12\log|t+3|}\]
Combine both integrals
We have:
\[I=I_1+I_2+C\]
Substituting the results:
\[I=\frac12\log|t+1|-\frac12\log|t+3|+C\]
Taking the common factor:
\[I=\frac12\left[\log|t+1|-\log|t+3|\right]+C\]
Using the logarithmic identity:
\[\log a-\log b=\log\left|\frac ab\right|\]
We obtain:
\[I=\frac12\log\left|\frac{t+1}{t+3}\right|+C\]
Substitute the original variable
Since:
\[t=x^2\]
Substituting back:
\[I=\frac12\log\left|\frac{x^2+1}{x^2+3}\right|+C\]
Therefore, the required integral is:
\[\boxed{\displaystyle\int\frac{2x}{(x^2+1)(x^2+3)}\,dx=\frac12\log\left(\frac{x^2+1}{x^2+3}\right)+C}\]

🎯 Exam Significance

Exam Significance

›

This problem is useful for CBSE Class 12 Mathematics board examination preparation because it demonstrates how substitution can simplify a rational integral before applying partial fraction decomposition.

Substitution: It tests the identification of an expression whose derivative appears in the numerator.
Partial fractions: It reinforces the decomposition of rational functions into simpler fractions.
Logarithmic integration: It tests the application of standard logarithmic integration formulae.
Algebraic accuracy: It develops the ability to compare coefficients and determine unknown constants.
Method selection: It demonstrates why choosing an appropriate substitution can make integration significantly simpler.

Significance for Competitive Entrance Examinations

The techniques applied in this problem are relevant to JEE Main, JEE Advanced, BITSAT and other competitive entrance examinations.

Recognition of derivatives: Students learn to identify \(2x\) as the derivative of \(x^2\).
Substitution techniques: It develops the ability to transform complicated rational integrals into simpler forms.
Partial fraction decomposition: It strengthens algebraic manipulation skills used in more advanced integration problems.
Logarithmic simplification: It reinforces the use of logarithmic identities.
Efficient problem-solving: It demonstrates how substitution can reduce lengthy calculations.

🚫 Common Mistakes

Common Mistakes

›

Incorrect partial fraction structure: Writing constant numerators directly over the original quadratic factors does not match the numerator \(2x\). Substitution is a more convenient approach here.
Incorrect differential: From \(t=x^2\), the correct differential is \(dt=2x\,dx\).
Missing substitution: Both quadratic expressions in the denominator must be converted into expressions involving t.
Incorrect partial fraction signs: The correct constants are \(A=\frac12\) and \(B=-\frac12\).
Incorrect logarithmic coefficients: Both logarithmic terms retain the factor \(\frac12\).
Incorrect substitution back: Replace t with \(x^2\) in both logarithmic terms.
Forgetting the integration constant: Always include C in the final indefinite integral.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

The numerator \(2x\) suggests the substitution \(t=x^2\).
The differential relationship is \(dt=2x\,dx\).
The transformed integral is \(\int dt/[(t+1)(t+3)]\).
The correct partial fraction decomposition is:
\[ \frac{1}{(t+1)(t+3)} = \frac{1}{2(t+1)} - \frac{1}{2(t+3)} \]
Both resulting terms are evaluated using logarithmic integration.
Recognising the derivative of an expression in the numerator can simplify a rational integral considerably.
Differentiation confirms the correctness of the final antiderivative.

📘 Concept & Theory

Theory and Concept

›

Integration Using Substitution and Partial Fraction Decomposition

This problem involves a rational function containing an even power of x in the denominator, together with a factor of x.

The expression \(x^4-1\) can be factorised using the difference of squares identity:

\[ \begin{aligned} x^4-1 &=(x^2)^2-1^2\\ &=(x^2-1)(x^2+1)\\ &=(x-1)(x+1)(x^2+1) \end{aligned} \]

However, a convenient approach is to use the substitution:

\[ t=x^2 \]

Differentiating:

\[ dt=2x\,dx \]

Therefore:

\[ \frac{dx}{x}=\frac{dt}{2t} \]

This transforms the original integral into a rational integral in t, which can be evaluated using partial fractions.

Important integration formulae:

\[ \int\frac{dt}{t}=\log|t|+C \]

\[ \int\frac{dt}{t-a}=\log|t-a|+C \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Substitute \(t=x^2\).
Obtain the differential relationship \(dx/x=dt/(2t)\).
Transform the original integral into a rational function of t.
Factorise \(t^2-1\).
Apply partial fraction decomposition.
Determine the unknown constants by comparing coefficients.
Integrate the resulting simple fractions.
Substitute \(t=x^2\) back into the expression.
Verify the final answer by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 37 steps

Given —
\[I=\int\frac{1}{x(x^4-1)}\,dx\]
Apply substitution
Let:
\[t=x^2\]
Differentiating both sides:
\[dt=2x\,dx\]
Dividing both sides by \(2x^2\):
\[\frac{dt}{2t}=\frac{dx}{x}\]
Thus:
\[\boxed{\frac{dx}{x}=\frac{dt}{2t}}\]
Transform the original integral
We have:
\[I=\int\frac{1}{x(x^4-1)}\,dx\]
Rearranging:
\[I=\int\frac{1}{x^4-1}\cdot\frac{dx}{x}\]
Since \(t=x^2\), we have \(x^4=t^2\). Substituting:
\[I=\int\frac{1}{t^2-1}\cdot\frac{dt}{2t}\]
Taking the constant factor outside:
\[\boxed{I=\frac12\int\frac{dt}{t(t^2-1)}}\]
Factorise the denominator
Using the difference of squares identity:
\[t^2-1=(t-1)(t+1)\]
Therefore:
\[I=\frac12\int\frac{dt}{t(t-1)(t+1)}\]
Apply partial fraction decomposition:
Assume:
\[\frac{1}{t(t-1)(t+1)}=\frac{A}{t}+\frac{B}{t-1}+\frac{C}{t+1}\]
Taking the LCM on the right-hand side:
\[ \begin{aligned} 1={}&A(t-1)(t+1)\\ &+Bt(t+1)+Ct(t-1) \end{aligned} \]
Expand the expression
Expanding each term:
\[1=A(t^2-1)+B(t^2+t)+C(t^2-t)\]
Collecting like terms:
\[1=(A+B+C)t^2+(B-C)t-A\]
Compare coefficients
Comparing the coefficients of \(t^2\):
\[A+B+C=0 \tag{1}\]
Comparing the coefficients of t:
\[B-C=0 \tag{2}\]
Comparing the constant terms:
\[-A=1 \tag{3}\]
From equation (3):
\[\boxed{A=-1}\]
From equation (2):
\[B=C\]
Substituting \(A=-1\) and \(C=B\) into equation (1):
\[-1+B+B=0\]
Simplifying:
\[2B=1\]
Therefore:
\[\boxed{B=\frac12}\]
Since \(C=B\):
\[\boxed{C=\frac12}\]
Hence, the required constants are:
\[\boxed{A=-1,\quad B=\frac12,\quad C=\frac12}\]
Write the partial fraction decomposition
Substituting the values of A, B and C:
\[\boxed{\frac{1}{t(t-1)(t+1)}=-\frac1t+\frac{1}{2(t-1)}+\frac{1}{2(t+1)}}\]
Therefore:
\[I=\frac12\int\left[-\frac1t+\frac{1}{2(t-1)}+\frac{1}{2(t+1)}\right]dt\]
Multiplying the factor \(1/2\) into each term:
\[I=-\frac12\int\frac{dt}{t}+\frac14\int\frac{dt}{t-1}+\frac14\int\frac{dt}{t+1}\]
Integrate each partial fraction
Using the standard logarithmic integration formulae:
\[\int\frac{dt}{t}=\log|t|+C\]

\[\int\frac{dt}{t-1}=\log|t-1|+C\]

\[\int\frac{dt}{t+1}=\log|t+1|+C\]
Therefore:
\[ \begin{aligned} I={}&-\frac12\log|t|\\ &+\frac14\log|t-1|\\ &+\frac14\log|t+1|+C \end{aligned} \]
Combine the logarithmic terms and taking \(\frac14\) common:
\[I=\frac14\left[-2\log|t|+\log|t-1|+\log|t+1|\right]+C\]
Using the logarithmic identities:
\[\log a+\log b=\log(ab)\]
and
\[2\log|t|=\log(t^2)\]
We obtain:
\[I=\frac14\log\left|\frac{(t-1)(t+1)}{t^2}\right|+C\]
Since \((t-1)(t+1)=t^2-1\):
\[\boxed{I=\frac14\log\left|\frac{t^2-1}{t^2}\right|+C}\]
Substitute the original variable
Since:
\[t=x^2\]
Therefore:
\[t^2=x^4\]
Substituting back:
\[I=\frac14\log\left|\frac{x^4-1}{x^4}\right|+C\]
Hence, the required integral is:
\[\boxed{\displaystyle\int\frac{1}{x(x^4-1)}\,dx=\frac14\log\left|\frac{x^4-1}{x^4}\right|+C}\]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int\frac{1}{x(x^4-1)}\,dx=\frac14\log\left|\frac{x^4-1}{x^4}\right|+C}\]

🎯 Exam Significance

Exam Significance

›

This problem is useful for CBSE Class 12 Mathematics board examination preparation because it combines substitution, factorisation and partial fraction decomposition.

Substitution: It tests the ability to recognise a suitable substitution involving an even power of x.
Factorisation: It reinforces the difference of squares identity.
Partial fractions: It develops the ability to decompose a rational function with three distinct linear factors.
Logarithmic integration: It tests the application of standard logarithmic integration formulae.
Algebraic accuracy: It requires careful determination of three unknown constants.
Verification: It demonstrates how differentiation can confirm an indefinite integral.

Significance for Competitive Entrance Examinations

The techniques used in this problem are relevant to JEE Main, JEE Advanced, BITSAT and other competitive entrance examinations.

Substitution techniques: Students learn to simplify rational functions involving higher powers of x.
Algebraic factorisation: Recognising \(x^4-1\) as a difference of squares makes the problem easier to solve.
Three-term partial fractions: The problem strengthens the ability to decompose rational functions involving three distinct linear factors after substitution.
Logarithmic simplification: It develops the ability to combine multiple logarithmic terms into a compact expression.
Derivative-based verification: It encourages checking the final result through differentiation.

🚫 Common Mistakes

Common Mistakes

›

Incorrect substitution: From \(t=x^2\), the correct differential relationship is \(dt=2x\,dx\), giving \(dx/x=dt/(2t)\).
Missing the factor of one-half: The transformed integral contains the factor \(\frac12\), which must be retained throughout the calculation.
Incorrect factorisation: Remember that \(t^2-1=(t-1)(t+1)\).
Incorrect partial fraction signs: The correct constants are \(A=-1\), \(B=\frac12\) and \(C=\frac12\).
Incorrect logarithmic coefficients: After multiplying by the external factor \(\frac12\), the coefficients become \(-\frac12,\frac14,\frac14\).
Incorrect substitution back: Since \(t=x^2\), the expression \(t^2\) becomes \(x^4\), not \(x^2\).
Ignoring absolute values: Retain absolute value signs in logarithmic expressions involving factors that may be negative.
Forgetting the integration constant: Always include C in the final indefinite integral.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

The substitution \(t=x^2\) simplifies the integral involving \(x^4-1\).
The differential relationship is \(dx/x=dt/(2t)\).
The denominator \(t^2-1\) factorises into \((t-1)(t+1)\).
The correct partial fraction decomposition is:
\[ \frac{1}{t(t^2-1)} = -\frac1t+\frac{1}{2(t-1)}+\frac{1}{2(t+1)} \]
The factor \(\frac12\) introduced during substitution must be applied to every partial fraction.
The resulting logarithmic terms can be combined into a single logarithm.
Differentiation confirms the correctness of the final antiderivative.

📘 Concept & Theory

Theory and Concept

›

Integration Using Exponential Substitution and Partial Fractions

The given integral contains an exponential expression \(e^x\) in the denominator. Since the derivative of \(e^x\) is also \(e^x\), we can simplify the integral by using the substitution:

\[ t=e^x \]

Differentiating both sides:

\[ dt=e^x\,dx \]

Since \(t=e^x\), we obtain:

\[ dx=\frac{dt}{t} \]

This transforms the original integral into a rational function of t:

\[ \int\frac{dt}{t(t-1)} \]

The resulting expression can be evaluated using partial fraction decomposition.

Important integration formulae:

\[ \int\frac{dt}{t}=\log|t|+C \]

\[ \int\frac{dt}{t-1}=\log|t-1|+C \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Substitute \(t=e^x\).
Differentiate to obtain \(dt=e^x\,dx\).
Express \(dx\) in terms of dt and t.
Transform the original integral into a rational integral.
Apply partial fraction decomposition.
Determine the unknown constants by comparing coefficients.
Integrate the resulting logarithmic terms.
Substitute \(t=e^x\) back into the result.
Verify the final answer by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 27 steps

Given —
\[I=\int\frac{1}{e^x-1}\,dx\]
Apply exponential substitution
Let:
\[t=e^x\]
Differentiating both sides with respect to x:
\[\frac{dt}{dx}=e^x\]
Therefore:
\[dt=e^x\,dx\]
Since \(e^x=t\):
\[dt=t\,dx\]
Rearranging:
\[\boxed{dx=\frac{dt}{t}}\]
Transform the original integral
We have:
\[I=\int\frac{1}{e^x-1}\,dx\]
Substituting \(e^x=t\) and \(dx=dt/t\):
\[I=\int\frac{1}{t-1}\cdot\frac{dt}{t}\]
Therefore:
\[\boxed{I=\int\frac{dt}{t(t-1)}}\]
Apply partial fraction decomposition
Assume:
\[\frac{1}{t(t-1)}=\frac{A}{t}+\frac{B}{t-1}\]
Taking the LCM on the right-hand side:
\[\frac{1}{t(t-1)}=\frac{A(t-1)+Bt}{t(t-1)}\]
Equating the numerators:
\[1=A(t-1)+Bt\]
Expand the expression
Expanding:
\[ \begin{aligned} 1&=At-A+Bt\\ &=(A+B)t-A \end{aligned} \]
Compare coefficients
Comparing the coefficients of t:
\[A+B=0 \tag{1}\]
Comparing the constant terms:
\[-A=1 \tag{2}\]
From equation (2):
\[\boxed{A=-1}\]
Substituting \(A=-1\) into equation (1):
\[-1+B=0\]
Therefore:
\[\boxed{B=1}\]
Hence, the required constants are:
\[\boxed{A=-1,\qquad B=1}\]
Write the partial fraction decomposition
Substituting the values of A and B:
\[\boxed{\frac{1}{t(t-1)}=-\frac1t+\frac1{t-1}}\]
Therefore:
\[I=-\int\frac{dt}{t}+\int\frac{dt}{t-1}\]
Integrate the partial fractions
Using the standard logarithmic integration formula:
\[\int\frac{dt}{t}=\log|t|+C\]
Similarly:
\[\int\frac{dt}{t-1}=\log|t-1|+C\]
Therefore:
\[ \begin{aligned} I &=-\log|t|+\log|t-1|+C\\ &=\log\left|\frac{t-1}{t}\right|+C \end{aligned} \]
Substitute the original variable
Since:
\[t=e^x\]
Substituting back:
\[I=\log\left|\frac{e^x-1}{e^x}\right|+C\]
Since \(e^x\) is always positive for real x, we can simplify:
\[ \begin{aligned} I &=\log|e^x-1|-\log(e^x)+C\\ &=\log|e^x-1|-x+C \end{aligned} \]
Therefore, the required integral is:
\[\boxed{\displaystyle\int\frac{1}{e^x-1}\,dx=\log|e^x-1|-x+C}\]

🎯 Exam Significance

Exam Significance

›

This problem is useful for CBSE Class 12 Mathematics board examination preparation because it combines exponential substitution, rational function integration and partial fraction decomposition.

Exponential substitution: It tests the recognition of an exponential expression whose derivative is proportional to itself.
Partial fractions: It reinforces the decomposition of rational functions into simpler fractions.
Logarithmic integration: It tests the application of standard logarithmic integration formulae.
Algebraic simplification: It develops the ability to combine logarithmic expressions correctly.
Verification: It demonstrates how differentiation can be used to validate an indefinite integral.

Significance for Competitive Entrance Examinations

The techniques applied in this problem are relevant to JEE Main, JEE Advanced, BITSAT and other competitive entrance examinations.

Exponential transformations: Students learn to convert exponential expressions into rational functions through substitution.
Partial fraction decomposition: It strengthens the ability to handle rational functions with distinct linear factors.
Logarithmic identities: It reinforces the simplification of logarithmic expressions.
Conceptual understanding: It demonstrates the relationship between exponential functions and logarithmic antiderivatives.
Efficient problem-solving: It shows how an appropriate substitution reduces a seemingly complicated integral to a standard form.

🚫 Common Mistakes

Common Mistakes

›

Incorrect differential: From \(t=e^x\), the correct relationship is \(dt=e^x\,dx\), giving \(dx=dt/t\).
Missing the factor t: Do not write \(dx=dt\). The correct expression is \(dx=dt/t\).
Incorrect partial fraction signs: The correct constants are \(A=-1\) and \(B=1\).
Incorrect logarithmic signs: The decomposition gives \(-\log|t|+\log|t-1|\).
Incorrect simplification: Since \(\log(e^x)=x\), the final expression becomes \(\log|e^x-1|-x+C\).
Forgetting absolute values: Retain absolute value signs around \(e^x-1\), which can be negative when \(x<0\).
Forgetting the integration constant: Always include C in the final indefinite integral.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

The substitution \(t=e^x\) converts the exponential expression into an algebraic variable.
The differential relationship is \(dx=dt/t\).
The transformed rational function is \(1/[t(t-1)]\).
The correct partial fraction decomposition is:
\[ \frac{1}{t(t-1)} = -\frac1t+\frac1{t-1} \]
The resulting integral is evaluated using logarithmic integration.
The identity \(\log(e^x)=x\) simplifies the final expression.
Differentiation confirms the correctness of the final antiderivative.

📘 Concept & Theory

Theory and Concept

›

This integral involves a rational function with two distinct linear factors in the denominator. We can evaluate it using partial fraction decomposition.

Assume:

\[ \frac{x}{(x-1)(x-2)} = \frac{A}{x-1}+\frac{B}{x-2} \]

✏️ Solution

Complete Solution

›

Step-by-step Solution · 11 steps

Apply partial fraction decomposition
Taking the LCM:
\[x=A(x-2)+B(x-1)\]
Expand the expression
\[ \begin{aligned} x&=Ax-2A+Bx-B\\ &=(A+B)x-(2A+B) \end{aligned} \]
Compare coefficients
Comparing the coefficients of x:
\[A+B=1 \tag{1}\]
Comparing the constant terms:
\[-2A-B=0 \tag{2}\]
From equation (1):
\[B=1-A\]
Substituting into equation (2):
\[ \begin{aligned} -2A-(1-A)&=0\\ -2A-1+A&=0\\ -A&=1\\ A&=-1 \end{aligned} \]
Therefore:
\[B=1-(-1)=2\]
Hence:
\[\boxed{A=-1,\quad B=2}\]
Substitute the constants
\[\frac{x}{(x-1)(x-2)}=-\frac1{x-1}+\frac2{x-2}\]
Integrate both terms
\[ \begin{aligned} I &=-\int\frac{dx}{x-1} +2\int\frac{dx}{x-2}\\ &=-\log|x-1|+2\log|x-2|+C \end{aligned} \]
Using logarithmic identities:
\[ \begin{aligned} I &=\log|(x-2)^2|-\log|x-1|+C\\ &=\log\left| \frac{(x-2)^2}{x-1} \right|+C \end{aligned} \]

💡 Answer

Final Answer

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Correct Option

Option (B), provided its intended expression is:

\[\boxed{\log\left|\frac{(x-2)^2}{x-1}\right|+C}\]

📘 Concept & Theory

Theory and Concept

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The given integrand is a rational function whose denominator contains a linear factor and an irreducible quadratic factor. We can simplify it using partial fraction decomposition.

Since the denominator is \(x(x^2+1)\), assume:

\[ \frac{1}{x(x^2+1)} = \frac{A}{x}+\frac{Bx+C}{x^2+1} \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Express the rational function in partial fractions.
Determine the unknown constants by comparing coefficients.
Integrate the resulting simpler expressions separately.
Match the result with the given options.
Verify the answer by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 12 steps

Apply partial fraction decomposition
\[\frac{1}{x(x^2+1)}=\frac{A}{x}+\frac{Bx+C}{x^2+1}\]
Taking the LCM and comparing numerators:
\[1=A(x^2+1)+x(Bx+C)\]
Expand the expression
\[ \begin{aligned} 1&=Ax^2+A+Bx^2+Cx\\ &= (A+B)x^2+Cx+A \end{aligned} \]
Compare coefficients
Comparing the coefficients of \(x^2\), \(x\), and the constant terms:
\[ \begin{aligned} A+B&=0\\ C&=0\\ A&=1 \end{aligned} \]
Therefore:
\[A=1,\quad B=-1,\quad C=0\]
Substituting these values into the partial fraction expression:
\[\boxed{\frac{1}{x(x^2+1)}=\frac1x-\frac{x}{x^2+1}}\]
Integrate both terms
\[ \begin{aligned} I &=\int\left(\frac1x-\frac{x}{x^2+1}\right)dx\\ &=\int\frac{dx}{x} -\int\frac{x}{x^2+1}\,dx \end{aligned} \]
For the second integral, let:
\[t=x^2+1\]
Differentiating:
\[dt=2x\,dx\quad\Rightarrow\quad x\,dx=\frac{dt}{2}\]
Therefore:
\[\begin{aligned}I&=\log|x|-\frac12\int\frac{dt}{t}\\ &=\log|x|-\frac12\log|t|+C\end{aligned}\]
Substituting \(t=x^2+1\):
\[I=\log|x|-\frac12\log|x^2+1|+C\]
Since \(x^2+1\) is always positive for real \(x\), the absolute value in its logarithm is unnecessary.
\[\boxed{\displaystyle\int\frac{dx}{x(x^2+1)}=\log|x|-\frac12\log(x^2+1)+C}\]

💡 Answer

Final Answer

›

Correct Option

Option (A)

\[\boxed{\log|x|-\frac12\log(x^2+1)+C}\]

🎯 Exam Significance

Exam Significance

›

Tests the application of partial fraction decomposition in integration.
Assesses the ability to integrate rational functions containing linear and irreducible quadratic factors.
Reinforces the standard logarithmic integration formula.
Tests accuracy in handling negative signs and numerical coefficients.

Significance for Competitive Entrance Examinations

Develops speed in identifying suitable partial fraction decompositions.
Tests recognition of integrals that can be simplified using substitution.
Strengthens the use of logarithmic identities in evaluating indefinite integrals.
Helps eliminate incorrect multiple-choice options through differentiation.

🚫 Common Mistakes

Common Mistakes

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Forgetting the factor \(\frac12\) when integrating \(\frac{x}{x^2+1}\).
Writing a positive sign instead of a negative sign before the second logarithmic term.
Ignoring the absolute value in \(\log|x|\).
Incorrectly assigning the numerator of the quadratic-factor fraction during partial fraction decomposition.

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 4 points

Use partial fractions when a rational function has a factorable denominator.
The useful decomposition here is:
\[ \frac{1}{x(x^2+1)} = \frac1x-\frac{x}{x^2+1} \]
Remember that:
\[ \int\frac{x}{x^2+a}\,dx = \frac12\log(x^2+a)+C \]
whenever the logarithm's argument is positive.
Differentiation provides an effective way to verify an indefinite integral.

Integrals

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Correct Option

Correct Option

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Integrals

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Correct Option

Correct Option

Significance for Competitive Entrance Examinations

Chapter Complete!

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Frequently Asked Questions

Q 01 What is the integral given in NCERT Class 12 Mathematics Integrals Exercise 7.5 Question 23?

Q 02 Which method is used to solve Exercise 7.5 Question 23?

Q 03 What is the partial fraction decomposition of 1/[x(x² + 1)]?

Q 04 What are the values of A, B and C in the partial fraction decomposition?

Q 05 What is the standard formula for integrating 1/x?

Q 06 How is the integral of x/(x² + 1) evaluated?

Q 07 What is the final answer to NCERT Integrals Exercise 7.5 Question 23?

Q 08 Which option is correct in Exercise 7.5 Question 23?

Q 09 How can the answer to this integration problem be verified?

Q 10 Why is partial fraction decomposition important for JEE and CBSE examinations?

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