NCERT Class 12 Maths Integrals Exercise 7.1 Solutions

📘 Concept & Theory

Theory and Concept

›

An antiderivative of a function \(f(x)\) is a function \(F(x)\) whose derivative is \(f(x)\). In mathematical notation,

\[ F'(x)=f(x) \]

Therefore, the indefinite integral of \(f(x)\) is written as

\[ \int f(x)\,dx=F(x)+C \]

Here, \(C\) is the arbitrary constant of integration. It is included because the derivative of every constant is zero. Thus, all functions differing by an additive constant have the same derivative.

Standard Integration Formulas

For solving this question, we use the following standard results.

Integration of a sine function
\[ \int \sin ax\,dx=-\frac{\cos ax}{a}+C,\qquad a\ne 0 \]
Integration of a cosine function
\[ \int \cos ax\,dx=\frac{\sin ax}{a}+C,\qquad a\ne 0 \]
Integration of an exponential function
\[ \int e^{ax}\,dx=\frac{e^{ax}}{a}+C,\qquad a\ne 0 \]
Integration of a power of a linear expression
\[ \int (ax+b)^n\,dx = \frac{(ax+b)^{n+1}}{a(n+1)}+C \]
where \(a\ne 0\) and \(n\ne -1\).
Linearity of integration
\[ \int [f(x)+g(x)]\,dx = \int f(x)\,dx+\int g(x)\,dx \]
and
\[ \int kf(x)\,dx=k\int f(x)\,dx \]
where \(k\) is a constant.

🗺️ Solution Roadmap

Step-by-step Plan

›

Recognise the standard derivative corresponding to the given function.
Account for the constant multiplying \(x\) inside the trigonometric, exponential, or algebraic expression.
Adjust the candidate antiderivative by dividing by the appropriate constant factor.
For a sum or difference of functions, integrate each term separately.
Add the arbitrary constant of integration \(C\).
Differentiate the final expression to verify that it reproduces the original integrand.

✏️ Solution

Q1. Find \(\displaystyle \int \sin 2x\,dx\)

›

Step-by-step Solution · 6 steps

Identify the relevant derivative
We know that the derivative of \(\cos u\), where \(u\) is a differentiable function of \(x\), is
\[\frac{d}{dx}(\cos u)=-\sin u\frac{du}{dx}\]
Taking \(u=2x\), we obtain
\[\frac{d}{dx}(\cos 2x)=-\sin 2x\frac{d}{dx}(2x)\]
Since \(\frac{d}{dx}(2x)=2\),
\[\frac{d}{dx}(\cos 2x)=-2\sin 2x\]
Adjust the constant factor.
The derivative of \(\cos 2x\) is \(-2\sin 2x\). To obtain \(\sin 2x\), we multiply \(\cos 2x\) by \(-\frac12\).
\[\frac{d}{dx}\left(-\frac12\cos 2x\right)=-\frac12(-2\sin 2x)=\sin 2x\]
Write the antiderivative.
Therefore,
\[\boxed{\int \sin 2x\,dx=-\frac12\cos 2x+C}\]
Verification
Differentiating the answer gives \(\sin 2x\), which is the original integrand. Hence, the result is correct.

✏️ Solution

Q2. Find \(\displaystyle \int \cos 3x\,dx\)

›

Step-by-step Solution · 7 steps

Identify the relevant derivative.
We know that
\[\frac{d}{dx}(\sin u)=\cos u\frac{du}{dx}\]
Taking \(u=3x\), we get
\[\frac{d}{dx}(\sin 3x)=\cos 3x\frac{d}{dx}(3x)\]
Since \(\frac{d}{dx}(3x)=3\),
\[\frac{d}{dx}(\sin 3x)=3\cos 3x\]
Adjust the constant factor.
To obtain \(\cos 3x\), divide \(\sin 3x\) by \(3\).
\[\frac{d}{dx}\left(\frac13\sin 3x\right)=\frac13(3\cos 3x)=\cos 3x\]
Write the antiderivative.
Therefore,
\[\boxed{\int \cos 3x\,dx=\frac13\sin 3x+C}\]
Verification
Differentiating \(\frac13\sin 3x+C\) gives \(\cos 3x\). Hence, the answer is correct.

✏️ Solution

Q3. Find \(\displaystyle \int e^{2x}\,dx\)

›

Step-by-step Solution · 5 steps

Recall the derivative of an exponential function.
For a constant \(a\),
\[\frac{d}{dx}(e^{ax})=ae^{ax}\]
Taking \(a=2\), we obtain
\[\frac{d}{dx}(e^{2x})=2e^{2x}\]
Adjust the constant factor.
The derivative of \(e^{2x}\) contains an extra factor of \(2\). Therefore, divide \(e^{2x}\) by \(2\).
\[\frac{d}{dx}\left(\frac12e^{2x}\right)=\frac12(2e^{2x})=e^{2x}\]
Write the antiderivative.
Thus,
\[\boxed{\int e^{2x}\,dx=\frac12e^{2x}+C}\]
Verification:
The derivative of \(\frac12e^{2x}+C\) is \(e^{2x}\), so the result is correct.

✏️ Solution

Q4. Find \(\displaystyle \int (ax+b)^2\,dx\)

›

Step-by-step Solution · 7 steps

Identify the algebraic pattern
The integrand is the square of a linear expression, \(ax+b\). We use the power rule in reverse and account for the derivative of the inner expression.

By the chain rule,
\[\frac{d}{dx}(ax+b)^3=3(ax+b)^2\frac{d}{dx}(ax+b)\]
Since \(a\) and \(b\) are constants,
\[\frac{d}{dx}(ax+b)=a\]
Therefore,
\[\frac{d}{dx}(ax+b)^3=3a(ax+b)^2\]
Adjust the constant factor.
To obtain \((ax+b)^2\), divide \((ax+b)^3\) by \(3a\), assuming \(a\ne 0\).
\[\frac{d}{dx}\left(\frac{(ax+b)^3}{3a}\right)=\frac{1}{3a}\left[3a(ax+b)^2\right]=(ax+b)^2\]
Write the antiderivative.
Hence, for \(a\ne 0\),
\[\boxed{\int (ax+b)^2\,dx=\frac{(ax+b)^3}{3a}+C}\]
Special Case
If \(a=0\), the integrand becomes the constant \(b^2\). In that case,
\[\int b^2\,dx=b^2x+C\]
Verification
For \(a\ne 0\), differentiating \(\frac{(ax+b)^3}{3a}+C\) gives \((ax+b)^2\), as required.

✏️ Solution

Q5. Find \(\displaystyle \int (\sin 2x-4e^{3x})\,dx\)

›

Step-by-step Solution · 7 steps

Apply the linearity of integration.
The integrand is the difference of two functions. We integrate each term separately while retaining its coefficient.
\[\begin{aligned}&\int(\sin 2x-4e^{3x})\,dx\\&\quad=\int\sin 2x\,dx-\int4e^{3x}\,dx\end{aligned}\]
Integrate the first term.
From part (i),
\[\int\sin 2x\,dx=-\frac12\cos 2x\]
Integrate the second term.
Since \(4\) is a constant, it can be taken outside the integral:
\[\int4e^{3x}\,dx=4\int e^{3x}\,dx\]
Using the exponential integration formula,
\[\int e^{3x}\,dx=\frac13e^{3x}\]
Therefore,
\[4\int e^{3x}\,dx=4\left(\frac13e^{3x}\right)=\frac43e^{3x}\]
Combine the results.
Substituting both antiderivatives, we obtain
\[\begin{aligned}&\int(\sin 2x-4e^{3x})\,dx\\&=-\frac12\cos 2x-\frac43e^{3x}+C\end{aligned}\]
Verification
Differentiating the result term by term gives
\[ \begin{aligned} &\frac{d}{dx}\left(-\frac12\cos 2x-\frac43e^{3x}+C\right)\\ &=-\frac12(-2\sin 2x)-\frac43(3e^{3x})\\ &=\sin 2x-4e^{3x} \end{aligned} \]
This is exactly the original integrand. Therefore, the solution is verified.

💡 Answer

Final Answer

›

🎯 Exam Significance

Exam Significance

›

This question develops the fundamental integration skills required for Class 12 Mathematics. Its main examination value lies in accurate application of standard formulas, correct handling of numerical coefficients, and systematic presentation of mathematical steps.

Formula recall: Students must remember the standard integrals of sine, cosine, and exponential functions.
Chain rule in reverse: Expressions such as \(\sin 2x\), \(e^{2x}\), and \((ax+b)^2\) require adjustment for the derivative of the inner function.
Algebraic accuracy: Correct signs and constant factors are essential for obtaining the right antiderivative.
Mathematical presentation: Writing the intermediate steps and the constant of integration helps demonstrate the method clearly.
Verification: Differentiating the final answer is an effective way to check its correctness.

Significance for Competitive Entrance Examinations

The techniques used here form the foundation for more advanced integration problems in JEE Main, JEE Advanced, and other entrance examinations. Although these examples are elementary, the same principles are used in more complex expressions involving substitutions, composite functions, and combinations of standard integrals.

Speed and accuracy: Recognising standard derivatives enables quick evaluation of straightforward indefinite integrals.
Reverse chain rule: The constant-factor adjustment becomes particularly important in composite functions.
Linearity: Separating sums and differences simplifies integrals containing multiple terms.
Error prevention: Remembering the negative sign in the integral of \(\sin ax\) and the denominator \(a\) in exponential integrals helps avoid common errors.
Foundation for advanced methods: These skills support later work involving substitution, integration by parts, partial fractions, and definite integrals.

🚫 Common Mistakes

Common Mistakes

›

Omitting the inner coefficient: Writing \(\int\sin 2x\,dx=-\cos 2x+C\) is incorrect. The correct coefficient is \(-\frac12\).
Using the wrong sign: The integral of \(\sin ax\) is negative cosine divided by \(a\), whereas the integral of \(\cos ax\) is positive sine divided by \(a\).
Forgetting the exponential denominator: The integral of \(e^{3x}\) is \(\frac13e^{3x}+C\), not \(e^{3x}+C\).
Ignoring the coefficient in a linear expression: For \(a\ne 0\), the integral of \((ax+b)^2\) contains the denominator \(3a\).
Dropping a coefficient: In \(\int4e^{3x}\,dx\), the coefficient \(4\) must remain, producing \(\frac43e^{3x}\).
Omitting the constant of integration: Every indefinite integral must include \(+C\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

Omitting the inner coefficient: Writing \(\int\sin 2x\,dx=-\cos 2x+C\) is incorrect. The correct coefficient is \(-\frac12\).
Using the wrong sign: The integral of \(\sin ax\) is negative cosine divided by \(a\), whereas the integral of \(\cos ax\) is positive sine divided by \(a\).
Forgetting the exponential denominator: The integral of \(e^{3x}\) is \(\frac13e^{3x}+C\), not \(e^{3x}+C\).
Ignoring the coefficient in a linear expression: For \(a\ne 0\), the integral of \((ax+b)^2\) contains the denominator \(3a\).
Dropping a coefficient: In \(\int4e^{3x}\,dx\), the coefficient \(4\) must remain, producing \(\frac43e^{3x}\).
Omitting the constant of integration: Every indefinite integral must include \(+C\).

📘 Concept & Theory

Theory and Concept

›

This problem uses the linearity of integration and the standard formula for integrating an exponential function. The integrand consists of two terms, \(4e^{3x}\) and \(1\), which can be integrated separately.

1. Sum rule of integration

The integral of a sum is the sum of the integrals:

\[ \int [f(x)+g(x)]\,dx = \int f(x)\,dx+\int g(x)\,dx \]

2. Constant multiple rule

A constant factor can be taken outside the integral:

\[ \int kf(x)\,dx=k\int f(x)\,dx \]

3. Exponential integration formula

For a nonzero constant \(a\),

\[ \int e^{ax}\,dx=\frac{e^{ax}}{a}+C \]

4. Integral of a constant

\[ \int 1\,dx=x+C \]

The constant of integration is arbitrary. When integrating multiple terms, we combine their constants into a single constant \(C\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Separate the integrand into the exponential term and the constant term.
Take the coefficient \(4\) outside the exponential integral.
Use the substitution \(t=3x\), so that \(dx=\frac{dt}{3}\), to integrate the exponential term.
Integrate the constant term \(1\) with respect to \(x\).
Substitute \(t=3x\) back into the result and combine the constants of integration.
Differentiate the final expression to verify the answer.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 16 steps

Apply the sum rule of integration.
Given —
\[I=\int(4e^{3x}+1)\,dx\]
Separating the two terms, we obtain
\[I=\int4e^{3x}\,dx+\int1\,dx\]
Take the constant coefficient outside the integral
Since \(4\) is a constant,
\[I=4\int e^{3x}\,dx+\int1\,dx\]
Evaluate the exponential integral using substitution.
Let
\[t=3x\]
Differentiating both sides with respect to \(x\),
\[\frac{dt}{dx}=3\]
Therefore,
\[dt=3\,dx\]
Rearranging,
\[dx=\frac{dt}{3}\]
Substituting \(t=3x\) and \(dx=\frac{dt}{3}\) into the exponential integral gives
\[\int e^{3x}\,dx=\int e^t\frac{dt}{3}\]
Taking the constant factor outside,
\[\int e^{3x}\,dx=\frac13\int e^t\,dt\]
Since the derivative of \(e^t\) with respect to \(t\) is \(e^t\),
\[\frac13\int e^t\,dt=\frac13e^t+C\]
Returning to the original variable \(x\), using \(t=3x\), we obtain
\[\int e^{3x}\,dx=\frac13e^{3x}+C\]
Multiplying by the coefficient \(4\),
\[4\int e^{3x}\,dx=\frac43e^{3x}+C\]
Integrate the constant term
We know that
\[\int1\,dx=x+C\]
Combine both results
Substituting the integrals of both terms into the original expression,
\[I=\frac43e^{3x}+x+C_1+C_2\]
Since the sum of two arbitrary constants is also an arbitrary constant, we write \(C_1+C_2=C\). Hence,
\[\boxed{\int(4e^{3x}+1)\,dx=\frac43e^{3x}+x+C}\]
Verification by Differentiation
To verify the result, differentiate the obtained antiderivative with respect to \(x\).

Let
\[F(x)=\frac43e^{3x}+x+C\]
Differentiating both sides,
\[ \begin{aligned} F'(x) &= \frac43\frac{d}{dx}(e^{3x}) +\frac{d}{dx}(x) +\frac{d}{dx}(C)\\[5pt] &= \frac43(3e^{3x})+1+0\\[5pt] &=4e^{3x}+1 \end{aligned} \]
The derivative matches the original integrand. Therefore, the answer is correct.

💡 Answer

Final Answer

›

\[\boxed{\int(4e^{3x}+1)\,dx=\frac43e^{3x}+x+C}\]

🎯 Exam Significance

Exam Significance

›

Application of standard formulas: Tests the ability to integrate exponential functions and constants.
Stepwise presentation: Demonstrates the sum rule, constant multiple rule, and substitution method clearly.
Chain rule in reverse: Reinforces the need to account for the coefficient \(3\) in the exponent \(3x\).
Constant of integration: Reinforces the correct use of \(C\) in indefinite integration.
Verification: Differentiation provides a reliable check on the final result.

Significance for Competitive Entrance Examinations

Quick evaluation: Recognising the standard exponential integral helps solve routine questions efficiently.
Substitution skills: The substitution \(t=3x\) is a basic example of the change-of-variable technique used in more complex integrals.
Coefficient accuracy: Correctly obtaining the factor \(\frac43\) is essential in objective questions where incorrect coefficients lead to wrong options.
Foundation for advanced integration: The same rules are used in definite integrals, differential equations, and applications of integration.

🚫 Common Mistakes

Common Mistakes

›

Forgetting to divide by \(3\) when integrating \(e^{3x}\).
Dropping the coefficient \(4\) from the exponential term.
Writing \(\int1\,dx=1\) instead of \(x+C\).
Forgetting the constant of integration in the final answer.
Failing to substitute \(t=3x\) back into the expression after integration.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

Integrate each term separately when the integrand is a sum.
Constant coefficients can be taken outside the integral.
For \(a\ne0\), \(\int e^{ax}\,dx=\frac{e^{ax}}{a}+C\).
The integral of \(1\) with respect to \(x\) is \(x+C\).
Always include the arbitrary constant in an indefinite integral.
Differentiate the final expression to verify the result.

📘 Concept & Theory

Theory and Concept

›

This problem illustrates the simplification of an algebraic expression before integration. The integrand contains a product of \(x^2\) and a bracketed expression involving a reciprocal power of \(x\). Expanding and simplifying the product reduces the integral to a difference of two elementary terms.

1. Algebraic distributive property

Multiply the factor outside the brackets by each term inside:

\[ a(b-c)=ab-ac \]

For the given expression,

\[ x^2\left(1-\frac{1}{x^2}\right) =x^2-\frac{x^2}{x^2} \]

Since \(\frac{x^2}{x^2}=1\) for \(x\ne0\), the integrand simplifies to

\[ x^2\left(1-\frac{1}{x^2}\right)=x^2-1 \]

2. Linearity of integration

The integral of a difference is the difference of the integrals:

\[ \int[f(x)-g(x)]\,dx = \int f(x)\,dx-\int g(x)\,dx \]

3. Power rule of integration

For any real number \(n\ne-1\),

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

In particular,

\[ \int x^2\,dx=\frac{x^3}{3}+C \]

Also,

\[ \int1\,dx=x+C \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Expand the bracket by multiplying \(x^2\) by each term.
Simplify the quotient \(\frac{x^2}{x^2}\), noting that the original expression is defined for \(x\ne0\).
Separate the resulting difference into two integrals.
Apply the power rule to \(\int x^2\,dx\) and integrate the constant term.
Combine the results and include the arbitrary constant of integration.
Verify the antiderivative by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 12 steps

Write the given integral
Let
\[I=\int x^2\left(1-\frac{1}{x^2}\right)\,dx\]
Expand the expression inside the integral
Using the distributive property,
\[I=\int\left(x^2\cdot1-x^2\cdot\frac{1}{x^2}\right)\,dx\]
Simplifying each product,
\[I=\int\left(x^2-\frac{x^2}{x^2}\right)\,dx\]
For \(x\ne0\), we have
\[\frac{x^2}{x^2}=1\]
Therefore,
\[I=\int(x^2-1)\,dx\]
Separate the integral
By the linearity of integration,
\[I=\int x^2\,dx-\int1\,dx\]
Integrate the first term
Applying the power rule with \(n=2\),
\[\int x^2\,dx=\frac{x^{2+1}}{2+1}=\frac{x^3}{3}\]
Integrate the second term
Since the integral of \(1\) with respect to \(x\) is \(x\),
\[\int1\,dx=x\]
Combine the results
Substituting the two antiderivatives,
\[I=\frac{x^3}{3}-x+C\]
Verification by Differentiation
Let
\[F(x)=\frac{x^3}{3}-x+C\]
Differentiating with respect to \(x\),
\[\begin{aligned}F'(x)&=\frac{d}{dx}\left(\frac{x^3}{3}-x+C\right)\\[5pt]&=\frac13(3x^2)-1+0\\[5pt]&=x^2-1\end{aligned}\]
From the original integrand, for \(x\ne0\),
\[x^2\left(1-\frac{1}{x^2}\right)=x^2-1\]
Thus, the derivative of the obtained antiderivative matches the original integrand throughout its domain. The answer is verified.

💡 Answer

Final Answer

›

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

🎯 Exam Significance

Exam Significance

›

Algebraic simplification: Demonstrates why simplifying the integrand before applying integration formulas can make a problem easier.
Application of standard formulas: Reinforces the power rule and integration of a constant.
Stepwise working: Shows how to present algebraic simplification and integration clearly.
Sign accuracy: Highlights the importance of retaining the minus sign when integrating a difference.
Domain awareness: The original expression contains \(\frac{1}{x^2}\), so it is undefined at \(x=0\). This is a useful reminder to consider the domain of the integrand.

Significance for Competitive Entrance Examinations

Efficient problem-solving: Simplifying the integrand first avoids unnecessary use of advanced integration methods.
Algebraic fluency: Correctly cancelling powers of \(x\) is essential in many integration and calculus problems.
Formula recognition: The problem reduces to elementary polynomial integration, which is frequently used as a step in more involved questions.
Accuracy under time constraints: Recognising simple cancellations can reduce calculation time and prevent avoidable errors.

🚫 Common Mistakes

Common Mistakes

›

Failing to distribute \(x^2\) across both terms inside the brackets.
Incorrectly simplifying \(\frac{x^2}{x^2}\); it equals \(1\) for \(x\ne0\).
Forgetting the negative sign in \(\int(x^2-1)\,dx\).
Writing \(\int x^2\,dx=x^2/2\) instead of applying the power rule correctly.
Omitting the arbitrary constant \(C\).
Ignoring that the original integrand is undefined at \(x=0\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

Simplify algebraic expressions before integrating whenever possible.
Use the distributive property to expand products involving brackets.
Apply the power rule to polynomial terms.
Integrate sums and differences term by term.
Include the arbitrary constant in every indefinite integral.
Check the final answer by differentiation.
The original integrand is defined only for \(x\ne0\), even though the simplified expression \(x^2-1\) is defined at \(x=0\).

📘 Concept & Theory

Theory and Concept

›

This problem illustrates the application of the linearity property of integration and the power rule to a polynomial of degree two. Since \(a\), \(b\), and \(c\) are constants, each term can be integrated separately, and its constant coefficient can be taken outside the integral.

1. Linearity property of integration

The integral of a sum is equal to the sum of the individual integrals:

\[ \int [f(x)+g(x)+h(x)]\,dx = \int f(x)\,dx+\int g(x)\,dx+\int h(x)\,dx \]

2. Constant multiple rule

If \(k\) is a constant, then

\[ \int kf(x)\,dx=k\int f(x)\,dx \]

3. Power rule of integration

For \(n\ne -1\),

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

Consequently, the standard results required for this question are

\[ \int x^2\,dx=\frac{x^3}{3}+C \]

\[ \int x\,dx=\frac{x^2}{2}+C \]

\[ \int 1\,dx=x+C \]

When these individual integrals are combined, their arbitrary constants can be represented by a single constant \(C\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given polynomial integral.
Separate the polynomial into three individual integrals using the linearity property.
Take the constant coefficients \(a\), \(b\), and \(c\) outside their respective integrals.
Apply the power rule to \(x^2\) and \(x\), and integrate the constant term.
Combine the results and include one arbitrary constant of integration.
Differentiate the final expression to verify the result.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 10 steps

Write the given integral
Let
\[I=\int (ax^2+bx+c)\,dx\]
Here, \(a\), \(b\), and \(c\) are constants.
Separate the integral term by term.
By the linearity property of integration,
\[I=\int ax^2\,dx+\int bx\,dx+\int c\,dx\]
Take the constant coefficients outside the integrals.
Since \(a\), \(b\), and \(c\) are constants, we obtain
\[I=a\int x^2\,dx+b\int x\,dx+c\int1\,dx\]
Integrate the first term.
Using the power rule with \(n=2\),
\[\int x^2\,dx=\frac{x^{2+1}}{2+1}=\frac{x^3}{3}\]
Therefore,
\[a\int x^2\,dx=\frac{a x^3}{3}\]
Integrate the second term.
Using the power rule with \(n=1\),
\[\int x\,dx=\frac{x^{1+1}}{1+1}=\frac{x^2}{2}\]
Therefore,
\[b\int x\,dx=\frac{b x^2}{2}\]
Integrate the constant term
Since \(c\) is a constant,
\[\int c\,dx=c\int1\,dx=cx\]
Combine the results
Substituting the three antiderivatives into the expression for \(I\),
\[I=\frac{a x^3}{3}+\frac{b x^2}{2}+cx+C\]
The constants of integration arising from the individual terms are combined into a single arbitrary constant \(C\).

💡 Answer

Final Answer

›

\[\boxed{\int (ax^2+bx+c)\,dx=\frac{a x^3}{3}+\frac{b x^2}{2}+cx+C}\]

🎯 Exam Significance

Exam Significance

›

Fundamental integration rules: This question tests the linearity property, constant multiple rule, and power rule.
Polynomial integration: It establishes the method for integrating polynomials term by term.
Stepwise presentation: Separating the terms and showing the integration of each term makes the solution clear and mathematically complete.
Constant of integration: Students should understand why the arbitrary constants from individual terms are combined into a single constant.
Verification: Differentiation confirms the correctness of the result and reinforces the inverse relationship between differentiation and integration.

Significance for Competitive Entrance Examinations

Speed and accuracy: Recognising a polynomial allows rapid evaluation without unnecessary substitutions.
Algebraic fluency: The method is useful when polynomial integration appears as one step in a longer calculus problem.
Generalisation: The same principle extends to polynomials of higher degree by integrating each power separately.
Foundation for advanced calculus: Polynomial antiderivatives are used in definite integrals, area calculations, and differential equations.
Parameter handling: Treating \(a\), \(b\), and \(c\) as constants develops the precision needed for parameter-based questions.

🚫 Common Mistakes

Common Mistakes

›

Forgetting that \(a\), \(b\), and \(c\) are constants with respect to \(x\).
Applying the power rule incorrectly, such as writing \(\int x^2\,dx=x^2/2\).
Forgetting to multiply the antiderivative of each term by its coefficient.
Omitting the term \(cx\) when integrating the constant \(c\).
Adding separate arbitrary constants unnecessarily instead of combining them into a single \(C\).
Forgetting the constant of integration in the final answer.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

Integrate a polynomial term by term using the linearity property.
Take constant coefficients outside their respective integrals.
Apply the power rule \(\int x^n\,dx=\frac{x^{n+1}}{n+1}+C\), for \(n\ne-1\).
The integral of a constant \(c\) is \(cx+C\).
Combine all arbitrary constants into one constant of integration.
Verify the result by differentiating the final expression.

📘 Concept & Theory

Theory and Concept

›

This problem applies the linearity property of integration, the constant multiple rule, the power rule, and the standard exponential integration formula. Since the integrand contains a polynomial term and an exponential term, we can integrate them separately and then combine the results.

1. Linearity property of integration

The integral of a sum is the sum of the individual integrals:

\[ \int [f(x)+g(x)]\,dx = \int f(x)\,dx+\int g(x)\,dx \]

2. Constant multiple rule

A constant factor can be taken outside the integral:

\[ \int kf(x)\,dx=k\int f(x)\,dx \]

3. Power rule of integration

For \(n\ne-1\),

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

In particular,

\[ \int x^2\,dx=\frac{x^3}{3}+C \]

4. Exponential integration formula

The derivative of \(e^x\) is \(e^x\). Therefore,

\[ \int e^x\,dx=e^x+C \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Separate the polynomial and exponential terms using the linearity property.
Take the constant coefficient \(2\) outside the polynomial integral.
Apply the power rule to integrate \(x^2\).
Integrate \(e^x\) using its standard formula.
Combine both antiderivatives and include one arbitrary constant of integration.
Differentiate the final expression to verify the result.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 7 steps

Write the given integral
Let
\[I=\int(2x^2+e^x)\,dx\]
Separate the terms
By the linearity property of integration,
\[I=\int2x^2\,dx+\int e^x\,dx\]
Take the constant coefficient outside the first integral
Since \(2\) is a constant,
\[I=2\int x^2\,dx+\int e^x\,dx\]
Integrate the polynomial term.
Applying the power rule with \(n=2\),
\[ \begin{aligned} \int x^2\,dx &=\frac{x^{2+1}}{2+1}\\[5pt] &=\frac{x^3}{3} \end{aligned} \]
Multiplying by \(2\), we obtain
\[2\int x^2\,dx=\frac{2x^3}{3}\]
Integrate the exponential term.
Using the standard formula,
\[\int e^x\,dx=e^x\]
Combine the results
Substituting both antiderivatives into the original expression gives
\[I=\frac{2x^3}{3}+e^x+C\]
The arbitrary constants arising from the individual integrals are combined into a single constant \(C\).

💡 Answer

Final Answer

›

\[\boxed{\int(2x^2+e^x)\,dx=\frac{2x^3}{3}+e^x+C}\]

🎯 Exam Significance

Exam Significance

›

Standard integration formulas: Reinforces the power rule and the integral of the exponential function \(e^x\).
Linearity of integration: Demonstrates how to integrate a sum by evaluating each term separately.
Correct coefficient handling: Tests the ability to retain the coefficient \(2\) throughout the calculation.
Mathematical presentation: Encourages clear, sequential working and correct use of the constant of integration.
Verification: Differentiation confirms that the final antiderivative is correct.

Significance for Competitive Entrance Examinations

Efficient calculations: Recognising standard integrals allows quick solutions to elementary integration questions.
Formula recall: The power rule and exponential formula are foundational tools for more advanced calculus problems.
Objective-question accuracy: Correct handling of coefficients and exponents helps avoid common errors in multiple-choice questions.
Foundation for advanced integration: These rules are used in definite integrals, differential equations, and other applications of calculus.

🚫 Common Mistakes

Common Mistakes

›

Forgetting to multiply the integral of \(x^2\) by \(2\).
Applying the power rule incorrectly to \(x^2\).
Writing \(\int e^x\,dx=xe^x+C\), which is incorrect.
Introducing an unnecessary denominator when integrating \(e^x\). The formula \(\frac{e^{ax}}{a}+C\) gives \(e^x+C\) when \(a=1\).
Omitting the constant of integration \(C\).
Failing to check the final expression by differentiation.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

Integrate sums term by term using the linearity property.
Take constant coefficients outside their respective integrals.
Use the power rule to integrate polynomial terms.
The exponential function \(e^x\) is its own derivative and antiderivative.
Combine arbitrary constants into one constant of integration.
Verify the antiderivative by differentiating the final result.

📘 Concept & Theory

Theory and Concept

›

This problem combines algebraic identities, simplification of rational expressions, the linearity property of integration, the power rule, and the standard integral of \(\frac{1}{x}\). The most efficient approach is to expand the square and simplify the integrand before integrating.

1. Algebraic identity

We use the identity

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

For the given expression, take

\[ a=\sqrt{x},\qquad b=\frac{1}{\sqrt{x}}. \]

Expanding the square gives

\[ \begin{aligned} \left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2 &=(\sqrt{x})^2 -2\left(\sqrt{x}\right)\left(\frac{1}{\sqrt{x}}\right) +\left(\frac{1}{\sqrt{x}}\right)^2\\[5pt] &=x-2+\frac{1}{x} \end{aligned} \]

2. Linearity property of integration

The integral of a sum or difference can be evaluated term by term:

\[ \int[f(x)+g(x)-h(x)]\,dx = \int f(x)\,dx+\int g(x)\,dx-\int h(x)\,dx \]

3. Power rule of integration

For \(n\ne -1\),

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

In particular,

\[ \int x\,dx=\frac{x^2}{2}+C \]

4. Integral of a constant

\[ \int k\,dx=kx+C \]

Therefore,

\[ \int 2\,dx=2x+C \]

5. Standard logarithmic integral

For \(x\ne0\),

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

Since the original integrand contains \(\sqrt{x}\) and \(\frac{1}{\sqrt{x}}\), its real-valued domain is \(x>0\). Hence, in this problem, \(\ln|x|=\ln x\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Identify the algebraic identity for the square of a difference.
Expand the square and simplify the products involving square roots.
Rewrite the integrand as a sum of \(x\), \(-2\), and \(\frac{1}{x}\).
Separate the integral into three simpler integrals.
Apply the power rule, integrate the constant term, and use the logarithmic integral formula.
Combine the results and include the constant of integration.
Differentiate the final expression to verify the answer.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 9 steps

Given —
\[I=\int\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2\,dx\]
Expand the square using the identity \((a-b)^2=a^2-2ab+b^2\).
\[\begin{aligned}I&=\int\left[(\sqrt{x})^2-2\left(\sqrt{x}\right)\left(\frac{1}{\sqrt{x}}\right)+\left(\frac{1}{\sqrt{x}}\right)^2\right]dx\end{aligned}\]
Simplify each term separately:
\[(\sqrt{x})^2=x\]

\[\left(\sqrt{x}\right)\left(\frac{1}{\sqrt{x}}\right)=1\]

\[\left(\frac{1}{\sqrt{x}}\right)^2=\frac{1}{x}\]
Substituting these results into the integral,
\[I=\int\left(x-2+\frac{1}{x}\right)\,dx\]
Separate the integral term by term
By the linearity property,
\[I=\int x\,dx-\int 2\,dx+\int\frac{1}{x}\,dx\]
Integrate the first term
Using the power rule with \(n=1\),
\[\begin{aligned}\int x\,dx&=\frac{x^{1+1}}{1+1}\\[5pt] &=\frac{x^2}{2}\end{aligned}\]
Integrate the second term
Since \(2\) is a constant,
\[\int 2\,dx=2x\]
The minus sign in the original expression must be retained. Thus, the contribution of this term is \(-2x\).
Integrate the third term.
Using the standard logarithmic integral formula,
\[\int\frac{1}{x}\,dx=\ln|x|\]
Because \(x>0\) in the original problem,
\[\ln|x|=\ln x\]
Therefore, the third term contributes \(\ln x\).
Combine all the results
Combining the three antiderivatives and including the arbitrary constant of integration \(C\), we obtain
\[I=\frac{x^2}{2}-2x+\ln x+C\]

💡 Answer

Final Answer

›

\[\boxed{\int\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2\,dx=\frac{x^2}{2}-2x+\ln x+C,\quad x>0}\]

🎯 Exam Significance

Exam Significance

›

Algebraic identities: Tests the ability to expand the square of a difference correctly.
Simplification before integration: Demonstrates how an expression involving radicals can be reduced to elementary terms.
Application of standard formulas: Reinforces polynomial integration, integration of constants, and the logarithmic integral.
Sign accuracy: Requires careful handling of the negative middle term and the subtraction of \(2x\).
Domain awareness: The original integrand is real-valued only for \(x>0\), which justifies writing \(\ln x\) instead of \(\ln|x|\).

Significance for Competitive Entrance Examinations

Efficient simplification: Expanding the square immediately reduces the problem to three standard integrals.
Formula recognition: Students must recognise when the logarithmic integral formula applies to a term of the form \(\frac{1}{x}\).
Algebraic accuracy: Simplifying powers and square roots correctly prevents errors in multi-step calculus questions.
Conceptual understanding: The problem connects algebraic identities with the power and logarithmic rules of integration.
Foundation for advanced problems: These techniques are useful in more involved indefinite and definite integrals.

🚫 Common Mistakes

Common Mistakes

›

Expanding \((a-b)^2\) incorrectly by omitting the middle term \(-2ab\).
Writing \(\left(\frac{1}{\sqrt{x}}\right)^2=\frac{1}{\sqrt{x}}\) instead of \(\frac{1}{x}\).
Forgetting that \(\sqrt{x}\cdot\frac{1}{\sqrt{x}}=1\) for \(x>0\).
Dropping the negative sign when integrating the constant term \(-2\).
Using \(\int\frac{1}{x}\,dx=\frac{x^0}{0}\), which is invalid because the power rule does not apply when the exponent is \(-1\).
Writing \(\ln x\) without considering the domain. Here, \(x>0\), so \(\ln x\) is appropriate.
Omitting the arbitrary constant of integration \(C\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

Use \((a-b)^2=a^2-2ab+b^2\) to expand the integrand.
Simplify radicals and reciprocal powers before integrating.
Apply the power rule to polynomial terms and integrate constants directly.
Remember the special formula \(\int\frac{1}{x}\,dx=\ln|x|+C\).
The original integrand requires \(x>0\), so the logarithmic term can be written as \(\ln x\).
Always include the arbitrary constant and verify the result by differentiation.

📘 Concept & Theory

Theory and Concept

›

This problem illustrates the simplification of a rational algebraic expression before integration. By dividing each term in the numerator by the denominator, we can rewrite the integrand as a sum of powers of \(x\). We can then apply the power rule, the logarithmic integral formula, and the linearity property of integration.

1. Simplification of algebraic fractions

When a polynomial is divided by a monomial, divide every term in the numerator by the denominator separately.

\[ \frac{x^3+5x-4}{x^2} = \frac{x^3}{x^2}+\frac{5x}{x^2}-\frac{4}{x^2} \]

Using the laws of exponents,

\[ \frac{x^m}{x^n}=x^{m-n},\qquad x\ne0 \]

we obtain

\[ \frac{x^3+5x-4}{x^2} = x+5x^{-1}-4x^{-2} \]

2. Linearity property of integration

The integral of a sum or difference can be evaluated term by term:

\[ \int[f(x)+g(x)-h(x)]\,dx = \int f(x)\,dx+\int g(x)\,dx-\int h(x)\,dx \]

3. Power rule of integration

For \(n\ne-1\),

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

This formula is applicable to \(x\) and \(x^{-2}\), but not to \(x^{-1}\).

4. Logarithmic integral formula

The special case \(n=-1\) is evaluated using

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

Since the original integrand is defined only for \(x\ne0\), the logarithmic expression must initially be written as \(\ln|x|\). On the positive domain \(x>0\), it can be written as \(\ln x\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Divide each term of the numerator by \(x^2\).
Rewrite the resulting expression using positive and negative powers of \(x\).
Separate the integral into three simpler integrals.
Apply the power rule to \(x\) and \(x^{-2}\).
Apply the logarithmic integral formula to \(x^{-1}\).
Combine the antiderivatives, preserve the correct signs, and add the constant of integration.
Differentiate the final expression to verify the result.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 11 steps

Given —
\[I=\int\frac{x^3+5x-4}{x^2}\,dx\]
Divide each numerator term by \(x^2\).
Using the distributive property of division,
\[I=\int\left(\frac{x^3}{x^2}+\frac{5x}{x^2}-\frac{4}{x^2}\right)\,dx\]
Simplifying each fraction separately,
\[\frac{x^3}{x^2}=x^{3-2}=x\]

\[\frac{5x}{x^2}=5x^{1-2}=\frac{5}{x}\]

\[\frac{4}{x^2}=4x^{-2}\]
Therefore,
\[I=\int\left(x+\frac{5}{x}-\frac{4}{x^2}\right)\,dx\]
Separate the integral term by term
Applying the linearity property,
\[I=\int x\,dx+5\int\frac{1}{x}\,dx-4\int x^{-2}\,dx\]
Integrate the first term
Using the power rule with \(n=1\),
\[\begin{aligned}\int x\,dx&=\frac{x^{1+1}}{1+1}\\[5pt] &=\frac{x^2}{2}\end{aligned}\]
Integrate the second term
We use the standard logarithmic integral formula:
\[\int\frac{1}{x}\,dx=\ln|x|\]
Multiplying by the coefficient \(5\),
\[5\int\frac{1}{x}\,dx=5\ln|x|\]
Integrate the third term.
Rewrite the reciprocal square as a negative power:
\[-4\int\frac{1}{x^2}\,dx=-4\int x^{-2}\,dx\]
Applying the power rule with \(n=-2\),
\[\begin{aligned}\int x^{-2}\,dx&=\frac{x^{-2+1}}{-2+1}\\[5pt] &=\frac{x^{-1}}{-1}\\[5pt] &=-x^{-1}\\[5pt] &=-\frac{1}{x}\end{aligned}\]
Multiplying by \(-4\), we obtain
\[-4\int x^{-2}\,dx=-4\left(-\frac{1}{x}\right)=\frac{4}{x}\]
Notice that the final term is positive because the negative coefficient \(-4\) multiplies the negative antiderivative \(-\frac1x\).
Combine all the results
Substituting the three antiderivatives into the original integral,
\[I=\frac{x^2}{2}+5\ln|x|+\frac{4}{x}+C\]

💡 Answer

Final Answer

›

\[\boxed{\int\frac{x^3+5x-4}{x^2}\,dx=\frac{x^2}{2}+5\ln|x|+\frac{4}{x}+C,\quad x\ne0}\]

For \(x>0\), the logarithmic term can equivalently be written as \(5\ln x\)

🎯 Exam Significance

Exam Significance

›

Algebraic simplification: Tests the ability to simplify rational expressions before integration.
Power rule: Reinforces integration of positive and negative integer powers of \(x\).
Logarithmic integral: Highlights the special case \(\int\frac1x\,dx=\ln|x|+C\), where the ordinary power rule cannot be applied.
Sign management: Requires careful handling of the negative coefficient in the term \(-\frac4{x^2}\).
Stepwise presentation: Demonstrates a clear method for simplifying the integrand, applying formulas, and verifying the answer.

Significance for Competitive Entrance Examinations

Time-efficient solving: Converting the rational expression into powers of \(x\) makes the integration straightforward.
Recognition of special cases: Distinguishing \(x^{-1}\) from other powers is essential for selecting the correct integration formula.
Accuracy in negative powers: The problem tests whether students can correctly integrate reciprocal powers without losing signs.
Foundation for advanced questions: The same simplification and integration techniques appear in more complex rational integrals.
Domain awareness: Recognising \(x\ne0\) helps students handle logarithmic terms and rational functions correctly.

🚫 Common Mistakes

Common Mistakes

›

Dividing only one term of the numerator by \(x^2\) instead of dividing every term.
Writing \(\frac{5x}{x^2}=5x\) instead of \(\frac5x\).
Applying the power rule to \(x^{-1}\), which would require division by zero.
Forgetting the absolute value in the general formula \(\int\frac1x\,dx=\ln|x|+C\).
Obtaining \(-\frac4x\) instead of \(+\frac4x\) after integrating \(-4x^{-2}\).
Omitting the arbitrary constant of integration \(C\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

Divide each term of the numerator by the denominator to simplify a rational integrand.
Rewrite reciprocal powers using negative exponents when convenient.
Apply the power rule for \(x^n\) when \(n\ne-1\).
Use \(\int\frac1x\,dx=\ln|x|+C\) for the special exponent \(n=-1\).
Pay close attention to signs when integrating negative powers.
Include the arbitrary constant and verify the result by differentiation.

📘 Concept & Theory

Theory and Concept

›

This problem demonstrates how to simplify an algebraic fraction involving a square root and then apply the power rule of integration. The numerator contains three terms, while the denominator is \(\sqrt{x}\). Dividing each term by the square root and expressing the resulting terms as fractional powers of \(x\) makes the integration straightforward.

1. Expressing square roots as fractional powers

We use the exponent identities

\[ \sqrt{x}=x^{1/2} \]

\[ \frac{1}{\sqrt{x}}=x^{-1/2} \]

For \(x>0\), the quotient of powers with the same base is simplified using

\[ \frac{x^m}{x^n}=x^{m-n} \]

Therefore,

\[ \frac{x^3}{\sqrt{x}} =x^{3-\frac12} =x^{5/2} \]

Similarly,

\[ \frac{3x}{\sqrt{x}} =3x^{1-\frac12} =3x^{1/2} \]

and

\[ \frac{4}{\sqrt{x}}=4x^{-1/2} \]

2. Linearity property of integration

The integral of a sum is the sum of the individual integrals:

\[ \int[f(x)+g(x)+h(x)]\,dx = \int f(x)\,dx+\int g(x)\,dx+\int h(x)\,dx \]

Constant coefficients can be taken outside their respective integrals:

\[ \int kf(x)\,dx=k\int f(x)\,dx \]

3. Power rule of integration

For \(n\ne-1\),

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

This formula applies to all three powers that occur in the simplified integrand: \(\frac52\), \(\frac12\), and \(-\frac12\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given integral and split the numerator into three separate terms.
Rewrite the square root in the denominator as \(x^{1/2}\).
Simplify each quotient using the law of exponents.
Separate the integral and take the constant coefficients outside.
Apply the power rule to each term, increasing each exponent by one and dividing by the new exponent.
Simplify the coefficients and express fractional powers in radical notation where appropriate.
Verify the final answer by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 14 steps

Given —
\[I=\int\frac{x^3+3x+4}{\sqrt{x}}\,dx\]
Separate the numerator into individual terms
Dividing each term in the numerator by \(\sqrt{x}\), we obtain
\[I=\int\left(\frac{x^3}{\sqrt{x}}+\frac{3x}{\sqrt{x}}+\frac{4}{\sqrt{x}}\right)\,dx\]
Express the terms using fractional exponents
Since \(\sqrt{x}=x^{1/2}\), we can rewrite the integral as
\[I=\int\left(\frac{x^3}{x^{1/2}}+\frac{3x}{x^{1/2}}+\frac{4}{x^{1/2}}\right)\,dx\]
Using the law of exponents, simplify the first term:
\[\frac{x^3}{x^{1/2}}=x^{3-\frac12}=x^{\frac{6}{2}-\frac12}=x^{5/2}\]
Simplify the second term:
\[\frac{3x}{x^{1/2}}=3x^{1-\frac12}=3x^{\frac12}\]
Simplify the third term:
\[\frac{4}{x^{1/2}}=4x^{-1/2}\]
Therefore, the integral becomes
\[I=\int\left(x^{5/2}+3x^{1/2}+4x^{-1/2}\right)\,dx\]
Separate the integral and take out the constant coefficients
By the linearity property,
\[I=\int x^{5/2}\,dx+3\int x^{1/2}\,dx+4\int x^{-1/2}\,dx\]
Integrate the first term
Applying the power rule with \(n=\frac52\),
\[\begin{aligned}\int x^{5/2}\,dx&=\frac{x^{\frac52+1}}{\frac52+1}\\[5pt] &=\frac{x^{7/2}}{\frac72}\\[5pt] &=\frac27x^{7/2}\end{aligned}\]
Integrate the second term
Applying the power rule with \(n=\frac12\),
\[\begin{aligned}\int x^{1/2}\,dx&=\frac{x^{\frac12+1}}{\frac12+1}\\[5pt] &=\frac{x^{3/2}}{\frac32}\\[5pt] &=\frac23x^{3/2}\end{aligned}\]
Multiplying by the coefficient \(3\),
\[\begin{aligned}3\int x^{1/2}\,dx&=3\left(\frac23x^{3/2}\right)\\[5pt] &=2x^{3/2}\end{aligned}\]
Integrate the third term
Applying the power rule with \(n=-\frac12\),
\[\begin{aligned}\int x^{-1/2}\,dx&=\frac{x^{-\frac12+1}}{-\frac12+1}\\[5pt] &=\frac{x^{1/2}}{\frac12}\\[5pt] &=2x^{1/2}\end{aligned}\]
Multiplying by the coefficient \(4\),
\[\begin{aligned}4\int x^{-1/2}\,dx&=4(2x^{1/2})\\[5pt] &=8x^{1/2}\\[5pt] &=8\sqrt{x}\end{aligned}\]
Combine the three antiderivatives
Substituting the results obtained in Steps 5, 6, and 7,
\[I=\frac27x^{7/2}+2x^{3/2}+8x^{1/2}+C\]
Expressing the final fractional power as a square root,
\[I=\frac27x^{7/2}+2x^{3/2}+8\sqrt{x}+C\]

💡 Answer

Final Answer

›

\[\boxed{\int\frac{x^3+3x+4}{\sqrt{x}}\,dx=\frac27x^{7/2}+2x^{3/2}+8\sqrt{x}+C,\quad x>0}\]

🎯 Exam Significance

Exam Significance

›

Fractional powers: Tests the ability to convert radicals into fractional exponents and apply exponent laws.
Power rule: Reinforces integration of positive and negative fractional powers.
Linearity: Demonstrates how to integrate a sum term by term.
Coefficient simplification: Requires careful arithmetic when dividing by fractional exponents.
Stepwise presentation: Shows the algebraic simplification and integration process clearly, supporting accurate mathematical communication.

Significance for Competitive Entrance Examinations

Rapid simplification: Converting the integrand into powers of \(x\) reduces the need for complicated methods.
Exponent manipulation: Correctly subtracting exponents is essential in algebraic and calculus-based objective questions.
Fractional arithmetic: The problem tests accurate handling of fractions such as \(\frac52+1=\frac72\).
Foundation for advanced integration: Fractional powers also occur in substitution-based integrals and more complex algebraic expressions.
Error prevention: Differentiation provides a quick way to check whether coefficients and exponents are correct.

🚫 Common Mistakes

Common Mistakes

›

Writing \(\frac{x^3}{\sqrt{x}}=x^{3/2}\) instead of \(x^{5/2}\).
Incorrectly simplifying \(\frac{3x}{\sqrt{x}}\); the correct result is \(3x^{1/2}\).
Changing the positive term \(\frac4{\sqrt{x}}\) into a negative term. All three terms in the original numerator are positive.
Forgetting that dividing by \(\frac72\) is equivalent to multiplying by \(\frac27\).
Calculating \(3\cdot\frac23\) incorrectly instead of obtaining \(2\).
Calculating \(4\cdot2\) incorrectly in the third term. The coefficient must be \(8\), not \(4\).
Omitting the constant of integration \(C\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

Rewrite square roots as fractional powers before integrating.
Use \(x^m/x^n=x^{m-n}\) to simplify algebraic fractions.
Apply the power rule to each term separately.
When the new exponent is a fraction, divide by that fraction carefully.
Preserve all coefficients and signs while simplifying the antiderivatives.
Include the arbitrary constant and verify the result by differentiation.

📘 Concept & Theory

Theory and Concept

›

This problem involves simplifying a rational algebraic expression before integration. The numerator is a polynomial that can be factorised by grouping its terms. By identifying the common factor \((x-1)\), we can simplify the fraction and reduce the integral to a polynomial that can be evaluated using the power rule.

1. Factorisation by grouping

When a polynomial contains several terms, group them strategically to identify a common factor.

For the numerator,

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

Taking the common factor from each group,

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

Taking \((x-1)\) as the common factor gives

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

2. Cancellation of a common factor

If the numerator and denominator contain the same nonzero factor, it can be cancelled. In this problem,

\[ \frac{(x-1)(x^2+1)}{x-1}=x^2+1, \qquad x\ne1 \]

The original integrand is undefined at \(x=1\), so the cancellation is valid only for \(x\ne1\).

3. Linearity property of integration

The integral of a sum is the sum of the individual integrals:

\[ \int[f(x)+g(x)]\,dx = \int f(x)\,dx+\int g(x)\,dx \]

4. Power rule of integration

For \(n\ne-1\),

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

In particular,

\[ \int x^2\,dx=\frac{x^3}{3}+C \]

Also, \(\int1\,dx=x+C\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given rational integral.
Group the numerator as \((x^3-x^2)+(x-1)\).
Factor out \((x-1)\) from both groups.
Cancel the common factor, noting that \(x\ne1\).
Separate the simplified integral into two terms.
Apply the power rule to \(x^2\) and integrate the constant term.
Combine the results and verify the answer by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 10 steps

Given —
\[I=\int\frac{x^3-x^2+x-1}{x-1}\,dx\]
Group the terms in the numerator
Group the first two terms together and the last two terms together:
\[x^3-x^2+x-1=(x^3-x^2)+(x-1)\]
Factorise each group
In the first group, \(x^2\) is the common factor:
\[x^3-x^2=x^2(x-1)\]
The second group already contains the factor \((x-1)\). Therefore,
\[x^3-x^2+x-1=x^2(x-1)+(x-1)\]
Taking \((x-1)\) common from the two terms,
\[x^3-x^2+x-1=(x-1)(x^2+1)\]
Substitute the factorised numerator
Substituting into the original integral gives
\[I=\int\frac{(x-1)(x^2+1)}{x-1}\,dx\]
Since the original denominator is nonzero only when \(x\ne1\), we cancel the common factor \((x-1)\):
\[I=\int(x^2+1)\,dx,\qquad x\ne1\]
Separate the integral
By the linearity property,
\[I=\int x^2\,dx+\int1\,dx\]
Integrate the first term
Using the power rule with \(n=2\),
\[\begin{aligned}\int x^2\,dx&=\frac{x^{2+1}}{2+1}\\[5pt] &=\frac{x^3}{3}\end{aligned}\]
Integrate the second term
Since the integral of \(1\) with respect to \(x\) is \(x\),
\[\int1\,dx=x\]
Combine the results
Combining the two antiderivatives and adding the arbitrary constant of integration \(C\),
\[I=\frac{x^3}{3}+x+C\]

💡 Answer

Final Answer

›

\[\boxed{\int\frac{x^3-x^2+x-1}{x-1}\,dx=\frac{x^3}{3}+x+C,\qquad x\ne1}\]

🎯 Exam Significance

Exam Significance

›

Factorisation by grouping: Tests the ability to recognise and factorise a polynomial before integration.
Rational expression simplification: Demonstrates how cancelling a common factor can reduce a complicated-looking fraction to a simple polynomial.
Application of standard formulas: Reinforces the power rule and integration of a constant.
Stepwise presentation: Shows the complete progression from factorisation to integration.
Domain awareness: Highlights that the original rational expression is undefined at \(x=1\), even though its simplified form is defined there.

Significance for Competitive Entrance Examinations

Efficient algebraic simplification: Factorisation by grouping can substantially reduce calculation time.
Pattern recognition: Recognising a common binomial factor is useful in algebra, calculus, and rational-function problems.
Accuracy: Correct cancellation avoids unnecessary integration techniques.
Conceptual understanding: The question connects polynomial factorisation with elementary integration.
Domain restrictions: Remembering the excluded value \(x=1\) develops mathematical precision when simplifying rational functions.

🚫 Common Mistakes

Common Mistakes

›

Grouping the numerator incorrectly and losing a sign.
Writing \(x^3-x^2=x(x-1)\) instead of \(x^2(x-1)\).
Failing to factorise the entire numerator as \((x-1)(x^2+1)\).
Cancelling \((x-1)\) without recognising that the original expression is undefined at \(x=1\).
Applying the power rule incorrectly to \(x^2\).
Omitting the arbitrary constant of integration \(C\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

Use factorisation by grouping to simplify suitable rational expressions.
Look for a common factor in both the numerator and denominator.
Simplify the integrand before applying integration formulas.
Use the power rule to integrate polynomial terms.
Retain domain restrictions from the original integrand, even after cancellation.
Include \(C\) and verify the final result by differentiation.

📘 Concept & Theory

Theory and Concept

›

This problem illustrates how to expand a product involving a square root, express the resulting terms as powers of \(x\), and apply the power rule of integration. The key is to simplify the integrand before integrating.

1. Distributive property

Multiply the factor \(\sqrt{x}\) by each term inside the bracket:

\[ (1-x)\sqrt{x}=\sqrt{x}-x\sqrt{x} \]

2. Expressing radicals as fractional powers

We use the identities

\[ \sqrt{x}=x^{1/2} \]

\[ x^m\cdot x^n=x^{m+n} \]

Therefore,

\[ x\sqrt{x}=x^1\cdot x^{1/2}=x^{3/2} \]

The integrand can consequently be written as

\[ (1-x)\sqrt{x}=x^{1/2}-x^{3/2} \]

3. Power rule of integration

For \(n\ne-1\),

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

Applying this rule to fractional powers gives

\[ \int x^{1/2}\,dx=\frac{x^{3/2}}{3/2} =\frac23x^{3/2}+C \]

and

\[ \int x^{3/2}\,dx=\frac{x^{5/2}}{5/2} =\frac25x^{5/2}+C \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given integral.
Expand the product \((1-x)\sqrt{x}\) using the distributive property.
Express the square root as \(x^{1/2}\) and simplify \(x\sqrt{x}\) to \(x^{3/2}\).
Separate the integral into two terms using linearity.
Apply the power rule to each fractional power.
Preserve the negative sign of the second term and add the constant of integration.
Differentiate the result to verify the answer.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 9 steps

Given —
\[I=\int(1-x)\sqrt{x}\,dx\]
Expand the product
Using the distributive property,
\[I=\int\left(\sqrt{x}-x\sqrt{x}\right)\,dx\]
Convert the radicals into fractional powers
Since \(\sqrt{x}=x^{1/2}\), we have
\[I=\int\left(x^{1/2}-x\cdot x^{1/2}\right)\,dx\]
Using the law of exponents,
\[x\cdot x^{1/2}=x^{1+1/2}=x^{3/2}\]
Therefore,
\[I=\int\left(x^{1/2}-x^{3/2}\right)\,dx\]
Separate the integral
By the linearity property of integration,
\[I=\int x^{1/2}\,dx-\int x^{3/2}\,dx\]
Integrate the first term
Applying the power rule with \(n=\frac12\),
\[ \begin{aligned} \int x^{1/2}\,dx &=\frac{x^{1/2+1}}{1/2+1}\\[5pt] &=\frac{x^{3/2}}{3/2}\\[5pt] &=\frac23x^{3/2} \end{aligned} \]
Integrate the second term
Applying the power rule with \(n=\frac32\),
\[ \begin{aligned} \int x^{3/2}\,dx &=\frac{x^{3/2+1}}{3/2+1}\\[5pt] &=\frac{x^{5/2}}{5/2}\\[5pt] &=\frac25x^{5/2} \end{aligned} \]
Since this integral is subtracted, its contribution to the final expression is
\[-\int x^{3/2}\,dx=-\frac25x^{5/2}\]
Combine the results
Combining both antiderivatives and adding the arbitrary constant \(C\), we obtain
\[I=\frac23x^{3/2}-\frac25x^{5/2}+C\]

💡 Answer

Final Answer

›

\[\boxed{\int(1-x)\sqrt{x}\,dx=\frac23x^{3/2}-\frac25x^{5/2}+C,\quad x\geq0}\]

🎯 Exam Significance

Exam Significance

›

Algebraic expansion: Tests the ability to expand a product before integration.
Fractional exponents: Reinforces the conversion of radicals into powers of \(x\).
Power rule: Demonstrates integration of positive fractional powers.
Sign accuracy: Requires the negative sign of the second term to be retained throughout the calculation.
Stepwise presentation: Encourages a clear progression from simplification to integration and verification.

Significance for Competitive Entrance Examinations

Efficient simplification: Expanding the product reduces the problem to two standard integrals.
Exponent manipulation: Correctly adding exponents when multiplying powers is an important algebraic skill.
Formula application: Reinforces the power rule for fractional exponents.
Calculation accuracy: Helps prevent mistakes involving fractional coefficients and negative signs.
Foundation for advanced integration: Similar expansions are useful in more complicated algebraic and definite integrals.

🚫 Common Mistakes

Common Mistakes

›

Failing to multiply both \(1\) and \(-x\) by \(\sqrt{x}\).
Writing \(x\sqrt{x}=x^{1/2}\) instead of \(x^{3/2}\).
Adding exponents incorrectly when simplifying \(x\cdot x^{1/2}\).
Dividing by the old exponent rather than the new exponent in the power rule.
Dropping the negative sign before the second antiderivative.
Omitting the constant of integration \(C\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

Expand products before integrating when this simplifies the integrand.
Write \(\sqrt{x}=x^{1/2}\) to apply the power rule conveniently.
Use \(x^m\cdot x^n=x^{m+n}\) to simplify products of powers.
Apply \(\int x^n\,dx=\frac{x^{n+1}}{n+1}+C\) for \(n\ne-1\).
Preserve all signs and coefficients when combining antiderivatives.
Verify the final result by differentiation.

📘 Concept & Theory

Theory and Concept

›

This problem applies the distributive property, laws of exponents, linearity of integration, and the power rule of integration. The integrand contains a square root multiplied by a polynomial. We simplify the product first and then integrate each term separately.

1. Expressing the square root as a fractional power

\[ \sqrt{x}=x^{1/2} \]

2. Laws of exponents

When powers with the same base are multiplied, their exponents are added:

\[ x^m\cdot x^n=x^{m+n} \]

Thus,

\[ x^2\sqrt{x}=x^{2+1/2}=x^{5/2} \]

\[ x\sqrt{x}=x^{1+1/2}=x^{3/2} \]

3. Linearity of integration

A sum or difference of integrable functions can be integrated term by term, and constant factors can be taken outside the integral.

\[ \int [af(x)+bg(x)]\,dx = a\int f(x)\,dx+b\int g(x)\,dx \]

4. Power rule of integration

For \(n\ne -1\),

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

This rule is applied to each fractional power obtained after expanding the integrand.

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given integral.
Express \(\sqrt{x}\) as \(x^{1/2}\).
Multiply \(x^{1/2}\) by each term of the polynomial and add the exponents.
Separate the integral into three simpler integrals.
Apply the power rule to each term, retaining the coefficients.
Combine the results and add the constant of integration.
Verify the result by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 9 steps

Given —
\[I=\int \sqrt{x}(3x^2+2x+3)\,dx\]
Express the square root as a fractional power
Since \(\sqrt{x}=x^{1/2}\),
\[I=\int x^{1/2}(3x^2+2x+3)\,dx\]
Expand the product
Multiplying \(x^{1/2}\) by each term inside the bracket gives
\[I=\int\left(3x^2x^{1/2}+2xx^{1/2}+3x^{1/2}\right)\,dx\]
Using the law \(x^m x^n=x^{m+n}\), simplify each product:
\[x^2x^{1/2}=x^{2+1/2}=x^{5/2}\]

\[xx^{1/2}=x^{1+1/2}=x^{3/2}\]
Therefore,
\[I=\int\left(3x^{5/2}+2x^{3/2}+3x^{1/2}\right)\,dx\]
Separate the integral
By the linearity property of integration,
\[I=3\int x^{5/2}\,dx+2\int x^{3/2}\,dx+3\int x^{1/2}\,dx\]
Integrate the first term
Applying the power rule with \(n=\frac52\),
\[ \begin{aligned} 3\int x^{5/2}\,dx &=3\left(\frac{x^{5/2+1}}{5/2+1}\right)\\[5pt] &=3\left(\frac{x^{7/2}}{7/2}\right)\\[5pt] &=3\cdot\frac27x^{7/2}\\[5pt] &=\frac67x^{7/2} \end{aligned} \]
Integrate the second term
Applying the power rule with \(n=\frac32\),
\[ \begin{aligned} 2\int x^{3/2}\,dx &=2\left(\frac{x^{3/2+1}}{3/2+1}\right)\\[5pt] &=2\left(\frac{x^{5/2}}{5/2}\right)\\[5pt] &=2\cdot\frac25x^{5/2}\\[5pt] &=\frac45x^{5/2} \end{aligned} \]
Integrate the third term.
Applying the power rule with \(n=\frac12\),
\[ \begin{aligned} 3\int x^{1/2}\,dx &=3\left(\frac{x^{1/2+1}}{1/2+1}\right)\\[5pt] &=3\left(\frac{x^{3/2}}{3/2}\right)\\[5pt] &=3\cdot\frac23x^{3/2}\\[5pt] &=2x^{3/2} \end{aligned} \]
Combine the results
Adding all three antiderivatives and including the arbitrary constant of integration \(C\), we obtain
\[I=\frac67x^{7/2}+\frac45x^{5/2}+2x^{3/2}+C\]

💡 Answer

Final Answer

›

\[\boxed{\int \sqrt{x}(3x^2+2x+3)\,dx=\frac67x^{7/2}+\frac45x^{5/2}+2x^{3/2}+C}\]

🎯 Exam Significance

Exam Significance

›

Algebraic simplification: Tests the ability to expand a product involving a radical and a polynomial.
Fractional powers: Reinforces the conversion of square roots into fractional exponents.
Power rule: Provides practice integrating powers such as \(x^{5/2}\), \(x^{3/2}\), and \(x^{1/2}\).
Stepwise working: Clear intermediate steps help demonstrate correct method and reduce avoidable errors.
Verification: Differentiation confirms that the antiderivative is correct.

Significance for Competitive Entrance Examinations

Exponent manipulation: Accurate use of exponent laws is essential for simplifying algebraic expressions.
Integration speed: Expanding first converts the problem into standard power-rule integrals.
Coefficient management: Taking constant factors outside the integrals reduces calculation complexity.
Error prevention: Keeping the coefficients \(3\), \(2\), and \(3\) attached to their respective terms prevents incorrect answers.
Transferable method: The same approach works for many integrals involving polynomials multiplied by fractional powers.

🚫 Common Mistakes

Common Mistakes

›

Forgetting to multiply every term of \(3x^2+2x+3\) by \(\sqrt{x}\).
Writing \(x^2\sqrt{x}=x^{3/2}\) instead of \(x^{5/2}\).
Adding the exponents incorrectly when simplifying \(x\sqrt{x}\).
Dividing by the original exponent rather than by the exponent increased by one.
Incorrectly simplifying \(\frac{3}{7/2}\), \(\frac{2}{5/2}\), or \(\frac{3}{3/2}\).
Omitting the constant of integration \(C\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

Convert radicals into fractional powers before applying the power rule.
Use the distributive property to expand the integrand.
Apply \(x^m x^n=x^{m+n}\) when multiplying powers with the same base.
Integrate each term separately using \(\int x^n\,dx=\frac{x^{n+1}}{n+1}+C\), for \(n\ne-1\).
Retain the coefficient of every term throughout the calculation.
Verify the result by differentiating the final expression.

📘 Concept & Theory

Theory and Concept

›

This problem uses the linearity property of integration and three standard integration formulas: the integral of a power of \(x\), the integral of cosine, and the integral of an exponential function. Since the integrand is a sum and difference of elementary functions, we can integrate each term separately.

1. Linearity property of integration

The integral of a sum or difference is the sum or difference of the individual integrals. Constant coefficients can be taken outside the integral.

\[ \int[af(x)+bg(x)]\,dx =a\int f(x)\,dx+b\int g(x)\,dx \]

2. Standard integration formulas

For a constant \(a\),

\[ \int x\,dx=\frac{x^2}{2}+C \]

\[ \int\cos x\,dx=\sin x+C \]

\[ \int e^x\,dx=e^x+C \]

The negative sign in the term \(-3\cos x\) must be retained when separating the integral. Therefore, its contribution is \(-3\sin x\), not \(+3\sin x\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given integral.
Separate the polynomial, trigonometric, and exponential terms using linearity.
Apply the standard integration formula to each term.
Preserve the negative sign of the cosine term.
Combine the antiderivatives and include the constant of integration.
Verify the result by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 6 steps

Fiven
\[I=\int(2x-3\cos x+e^x)\,dx\]
Separate the integral
By the linearity property of integration,
\[I=\int2x\,dx-3\int\cos x\,dx+\int e^x\,dx\]
Integrate the first term
Taking the constant \(2\) outside the integral and applying the power rule,
\[ \begin{aligned} \int2x\,dx &=2\int x\,dx\\[5pt] &=2\left(\frac{x^2}{2}\right)\\[5pt] &=x^2 \end{aligned} \]
Integrate the second term.
[Using \(\int\cos x\,dx=\sin x\),
\[\begin{aligned}-3\int\cos x\,dx&=-3\sin x\end{aligned}\]
The negative sign remains because the original integrand contains \(-3\cos x\).
Integrate the third term
Since the exponential function \(e^x\) is its own derivative,
\[\int e^x\,dx=e^x\]
Combine the results
Adding the three antiderivatives and including the arbitrary constant \(C\), we obtain
\[I=x^2-3\sin x+e^x+C\]

💡 Answer

Final Answer

›

\[ \boxed{\int(2x-3\cos x+e^x)\,dx =x^2-3\sin x+e^x+C} \]

🎯 Exam Significance

Exam Significance

›

Standard formulas: Tests knowledge of basic polynomial, trigonometric, and exponential integrals.
Linearity: Reinforces the method of splitting a sum or difference into separate integrals.
Sign accuracy: Highlights the importance of retaining negative signs while integrating trigonometric terms.
Presentation: Writing each integration step clearly helps demonstrate the method and avoid avoidable errors.
Verification: Differentiation provides a reliable way to check an indefinite integral.

Significance for Competitive Entrance Examinations

Rapid evaluation: Recognising standard integrals allows quick solution of direct questions.
Mixed-function practice: Builds fluency with algebraic, trigonometric, and exponential functions in a single expression.
Sign management: Prevents common errors involving negative coefficients.
Foundation for advanced problems: These formulas are frequently used alongside substitution, integration by parts, and other integration techniques.

🚫 Common Mistakes

Common Mistakes

›

Writing \(+3\sin x\) instead of \(-3\sin x\).
Forgetting to multiply the integral of \(x\) by the coefficient \(2\).
Incorrectly integrating \(\cos x\) as \(-\sin x\). In fact, \(\int\cos x\,dx=\sin x+C\).
Changing \(\int e^x\,dx\) into an expression involving an unnecessary denominator.
Omitting the constant of integration \(C\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

Integrate sums and differences term by term.
Use \(\int x\,dx=\frac{x^2}{2}+C\).
Remember that \(\int\cos x\,dx=\sin x+C\).
The exponential function satisfies \(\int e^x\,dx=e^x+C\).
Preserve the original signs and coefficients throughout the calculation.
Differentiate the final expression to verify the answer.

📘 Concept & Theory

Theory and Concept

›

This problem combines three standard integration techniques: the power rule for polynomial terms, the standard integral of the sine function, and the power rule applied to a square root. The integrand is a sum and difference of separate functions, so the integral can be evaluated term by term.

1. Linearity property of integration

The integral of a sum or difference is the sum or difference of the individual integrals. Constant coefficients can be taken outside the integral.

\[ \int[af(x)+bg(x)]\,dx =a\int f(x)\,dx+b\int g(x)\,dx \]

2. Power rule of integration

For \(n\ne -1\),

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

3. Standard trigonometric integral

\[ \int\sin x\,dx=-\cos x+C \]

Therefore, the integral of \(-3\sin x\) is \(+3\cos x\), because the two negative signs cancel.

4. Expressing a square root as a fractional power

\[ \sqrt{x}=x^{1/2} \]

Consequently,

\[ \int\sqrt{x}\,dx =\int x^{1/2}\,dx =\frac{x^{3/2}}{3/2} =\frac23x^{3/2}+C \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given integral.
Separate the three terms using the linearity property.
Apply the power rule to \(2x^2\).
Integrate \(-3\sin x\), paying attention to the negative signs.
Convert \(\sqrt{x}\) into \(x^{1/2}\) and apply the power rule.
Combine the results and add the constant of integration.
Verify the answer by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 7 steps

Given —
\[I=\int\left(2x^2-3\sin x+5\sqrt{x}\right)\,dx\]
Separate the integral
Using the linearity property,
\[ I=2\int x^2\,dx-3\int\sin x\,dx +5\int\sqrt{x}\,dx \]
Integrate the polynomial term
Applying the power rule with \(n=2\),
\[ \begin{aligned} 2\int x^2\,dx &=2\left(\frac{x^{2+1}}{2+1}\right)\\[5pt] &=2\left(\frac{x^3}{3}\right)\\[5pt] &=\frac23x^3 \end{aligned} \]
Integrate the trigonometric term
Since
\[\int\sin x\,dx=-\cos x\]
we obtain
\[ \begin{aligned} -3\int\sin x\,dx &=-3(-\cos x)\\[5pt] &=3\cos x \end{aligned} \]
Notice that the final sign is positive. This is a frequent source of errors in trigonometric integration.
Integrate the square-root term
First, express the square root as a fractional power:
\[\sqrt{x}=x^{1/2}\]
Therefore,
\[ \begin{aligned} 5\int\sqrt{x}\,dx &=5\int x^{1/2}\,dx\\[5pt] &=5\left(\frac{x^{1/2+1}}{1/2+1}\right)\\[5pt] &=5\left(\frac{x^{3/2}}{3/2}\right)\\[5pt] &=5\cdot\frac23x^{3/2}\\[5pt] &=\frac{10}{3}x^{3/2} \end{aligned} \]
Combine the results.
Adding the three antiderivatives and including the arbitrary constant of integration \(C\),
\[I=\frac23x^3+3\cos x+\frac{10}{3}x^{3/2}+C\]

💡 Answer

Final Answer

›

\[ \boxed{ \int\left(2x^2-3\sin x+5\sqrt{x}\right)\,dx =\frac23x^3+3\cos x+\frac{10}{3}x^{3/2}+C } \]

🎯 Exam Significance

Exam Significance

›

Application of standard formulas: Tests the use of the power rule and basic trigonometric integration.
Radical expressions: Reinforces converting square roots into fractional powers before integrating.
Sign accuracy: Tests the correct handling of the negative coefficient in \(-3\sin x\).
Stepwise presentation: Clear intermediate calculations make the method easy to follow and check.
Verification: Differentiating the final answer confirms its correctness.

Significance for Competitive Entrance Examinations

Formula recall: Builds fluency with standard polynomial and trigonometric integrals.
Calculation accuracy: Encourages careful simplification of numerical coefficients and fractional exponents.
Mixed integrands: Provides practice integrating algebraic and trigonometric terms together.
Time efficiency: Recognising the linearity property helps solve similar direct-integration questions quickly.
Foundation for advanced methods: These basic rules are frequently used within substitution and integration by parts problems.

🚫 Common Mistakes

Common Mistakes

›

Writing \(-3\cos x\) instead of \(+3\cos x\) after integrating \(-3\sin x\).
Forgetting to multiply the result of each integral by its original coefficient.
Integrating \(x^{1/2}\) as \(x^{1/2}/(1/2)\) without first increasing the exponent by one.
Simplifying \(5\cdot\frac23\) incorrectly; the correct coefficient is \(\frac{10}{3}\).
Omitting the constant of integration \(C\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

Use the linearity property to integrate each term separately.
Apply the power rule \(\int x^n\,dx=\frac{x^{n+1}}{n+1}+C\), for \(n\ne-1\).
Remember that \(\int\sin x\,dx=-\cos x+C\).
Convert \(\sqrt{x}\) to \(x^{1/2}\) before applying the power rule.
Track negative signs and numerical coefficients carefully.
Verify an indefinite integral by differentiating the result.

📘 Concept & Theory

Theory and Concept

›

This problem uses the distributive property of multiplication, the linearity property of integration, and two standard trigonometric integration formulas. The key is to expand the product and recognise the derivatives of \(\tan x\) and \(\sec x\).

1. Distributive property

Multiply \(\sec x\) by each term inside the bracket:

\[ \sec x(\sec x+\tan x) =\sec^2x+\sec x\tan x \]

2. Linearity property of integration

\[ \int[f(x)+g(x)]\,dx =\int f(x)\,dx+\int g(x)\,dx \]

3. Standard trigonometric integrals

Since

\[ \frac{d}{dx}(\tan x)=\sec^2x \]

and

\[ \frac{d}{dx}(\sec x)=\sec x\tan x \]

their corresponding integrals are

\[ \int\sec^2x\,dx=\tan x+C \]

\[ \int\sec x\tan x\,dx=\sec x+C \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given integral.
Expand the product using the distributive property.
Separate the expression into two integrals.
Apply the standard integrals of \(\sec^2x\) and \(\sec x\tan x\).
Combine the results and add a single arbitrary constant \(C\).
Verify the result by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 6 steps

Given —
\[I=\int\sec x(\sec x+\tan x)\,dx\]
Expand the product
Applying the distributive property,
\[ \begin{aligned} \sec x(\sec x+\tan x) &=\sec x\cdot\sec x+\sec x\cdot\tan x\\ &=\sec^2x+\sec x\tan x \end{aligned} \]
Therefore,
\[I=\int(\sec^2x+\sec x\tan x)\,dx\]
Separate the integral
Using the linearity property,
\[I=\int\sec^2x\,dx+\int\sec x\tan x\,dx\]
Integrate the first term
Since the derivative of \(\tan x\) is \(\sec^2x\),
\[\int\sec^2x\,dx=\tan x\]
Integrate the second term
Since the derivative of \(\sec x\) is \(\sec x\tan x\),
\[\int\sec x\tan x\,dx=\sec x\]
Combine the results
Adding both antiderivatives and including the arbitrary constant of integration,
\[I=\tan x+\sec x+C\]

💡 Answer

Final Answer

›

\[\boxed{\int\sec x(\sec x+\tan x)\,dx=\tan x+\sec x+C}\]

🎯 Exam Significance

Exam Significance

›

Standard formulas: Tests the direct application of two important trigonometric integration formulas.
Algebraic manipulation: Requires correct expansion of a product before integration.
Recognition of derivatives: Reinforces the relationship between differentiation and integration.
Stepwise presentation: Demonstrates a clear method that is easy to follow and verify.
Constant of integration: Reinforces the need to include \(C\) in an indefinite integral.

Significance for Competitive Entrance Examinations

Quick recognition: Identifying standard derivative patterns allows rapid evaluation.
Trigonometric fluency: Builds familiarity with expressions involving secant and tangent.
Algebraic efficiency: Expanding the product converts the question into two standard integrals.
Error control: Differentiation provides an efficient check of the final result.
Foundation for advanced integration: Recognising derivative patterns is useful in substitution and other integration techniques.

🚫 Common Mistakes

Common Mistakes

›

Failing to expand \(\sec x(\sec x+\tan x)\) correctly.
Confusing \(\int\sec^2x\,dx=\tan x+C\) with the integral of \(\tan^2x\).
Writing \(\int\sec x\tan x\,dx=\tan x+C\) instead of \(\sec x+C\).
Adding separate constants after each integral without simplifying them into one arbitrary constant.
Forgetting the constant of integration \(C\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

Expand products before integrating when doing so simplifies the integrand.
Remember that \(\int\sec^2x\,dx=\tan x+C\).
Remember that \(\int\sec x\tan x\,dx=\sec x+C\).
Use the linearity property to integrate sums term by term.
Verify the result by differentiating the final expression.

📘 Concept & Theory

Theory and Concept

›

This problem uses reciprocal trigonometric identities, the quotient of trigonometric functions, the Pythagorean identity, and standard integration formulas. The key strategy is to simplify the given expression into a form involving \(\tan^2x\), which can be integrated using a trigonometric identity.

1. Reciprocal trigonometric identities

\[ \sec x=\frac{1}{\cos x}, \qquad \operatorname{cosec}x=\frac{1}{\sin x} \]

Consequently,

\[ \sec^2x=\frac{1}{\cos^2x}, \qquad \operatorname{cosec}^2x=\frac{1}{\sin^2x} \]

2. Quotient identity

\[ \tan x=\frac{\sin x}{\cos x} \]

Therefore,

\[ \frac{\sec^2x}{\operatorname{cosec}^2x} =\frac{1/\cos^2x}{1/\sin^2x} =\frac{\sin^2x}{\cos^2x} =\tan^2x \]

3. Pythagorean trigonometric identity

The fundamental identity is

\[ 1+\tan^2x=\sec^2x \]

Rearranging gives

\[ \tan^2x=\sec^2x-1 \]

4. Standard integration formulas

\[ \int\sec^2x\,dx=\tan x+C \]

\[ \int 1\,dx=x+C \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given integral.
Replace \(\sec^2x\) and \(\operatorname{cosec}^2x\) with their reciprocal forms.
Simplify the quotient to obtain \(\tan^2x\).
Use \(\tan^2x=\sec^2x-1\) to rewrite the integrand.
Separate the integral into two standard integrals.
Integrate each term and combine the results.
Verify the answer by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 8 steps

Given —
\[I=\int\frac{\sec^2x}{\operatorname{cosec}^2x}\,dx\]
Apply the reciprocal identities
Using \(\sec^2x=\frac{1}{\cos^2x}\) and \(\operatorname{cosec}^2x=\frac{1}{\sin^2x}\),
We obtain
\[I=\int\frac{1/\cos^2x}{1/\sin^2x}\,dx\]
Dividing by a fraction is equivalent to multiplying by its reciprocal. Hence,
\[ \begin{aligned} I &=\int\frac{1}{\cos^2x}\cdot\sin^2x\,dx\\ &=\int\frac{\sin^2x}{\cos^2x}\,dx\\ &=\int\tan^2x\,dx \end{aligned} \]
Use the Pythagorean identity
From the identity
\[1+\tan^2x=\sec^2x\]
we get
\[\tan^2x=\sec^2x-1\]
Substituting this into the integral,
\[I=\int(\sec^2x-1)\,dx\]
Separate the integral
Applying the linearity property,
\[I=\int\sec^2x\,dx-\int1\,dx\]
Integrate each term
Since
\[\int\sec^2x\,dx=\tan x\]
and
\[\int1\,dx=x\]
the integral becomes
\[I=\tan x-x+C\]

💡 Answer

Final Answer

›

\[\boxed{\int\frac{\sec^2x}{\operatorname{cosec}^2x}\,dx=\tan x-x+C}\]

🎯 Exam Significance

Exam Significance

›

Trigonometric identities: Tests the application of reciprocal, quotient, and Pythagorean identities.
Algebraic simplification: Demonstrates how simplifying the integrand can make an apparently complicated integral straightforward.
Standard integration: Reinforces the integral of \(\sec^2x\) and the integral of a constant.
Logical presentation: Shows the importance of presenting each identity and simplification clearly.
Verification: Differentiation confirms the correctness of the antiderivative.

Significance for Competitive Entrance Examinations

Identity recognition: Recognising \(\tan^2x=\sec^2x-1\) avoids unnecessary integration techniques.
Time efficiency: Simplifying before integrating reduces calculation effort.
Trigonometric fluency: Strengthens the ability to move between reciprocal and quotient forms.
Conceptual understanding: Connects trigonometric identities with standard derivatives and antiderivatives.
Error prevention: Simplification makes it easier to identify sign errors and incorrect reciprocal operations.

🚫 Common Mistakes

Common Mistakes

›

Incorrectly simplifying the quotient of \(\sec^2x\) and \(\operatorname{cosec}^2x\). Remember that dividing by \(1/\sin^2x\) means multiplying by \(\sin^2x\).
Writing \(\tan^2x=\sec^2x+1\) instead of \(\tan^2x=\sec^2x-1\).
Forgetting the negative sign when integrating the constant term \(-1\).
Writing \(\int\sec^2x\,dx=\sec x+C\) instead of \(\tan x+C\).
Omitting the constant of integration \(C\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

Use reciprocal identities to simplify expressions involving \(\sec x\) and \(\operatorname{cosec}x\).
Recognise that \(\frac{\sec^2x}{\operatorname{cosec}^2x}=\tan^2x\).
Convert \(\tan^2x\) into \(\sec^2x-1\) using a Pythagorean identity.
Apply standard integration formulas after simplifying the integrand.
Check the final answer by differentiation.

📘 Concept & Theory

Theory and Concept

›

This problem applies algebraic splitting, reciprocal and quotient trigonometric identities, the linearity property of integration, and standard trigonometric integration formulas. The most efficient strategy is to split the numerator into separate terms and express the resulting fractions in terms of \(\sec x\) and \(\tan x\).

1. Splitting a fraction

Using the property

\[ \frac{a-b}{c}=\frac{a}{c}-\frac{b}{c}, \]

we can write

\[ \frac{2-3\sin x}{\cos^2x} =\frac{2}{\cos^2x}-\frac{3\sin x}{\cos^2x}. \]

2. Reciprocal and quotient identities

Recall that

\[ \sec x=\frac{1}{\cos x}, \qquad \tan x=\frac{\sin x}{\cos x}. \]

Therefore,

\[ \frac{2}{\cos^2x}=2\sec^2x \]

and

\[ \frac{3\sin x}{\cos^2x}=3\sec x\tan x. \]

3. Standard trigonometric integration formulas

\[ \int\sec^2x\,dx=\tan x+C \]

\[ \int\sec x\tan x\,dx=\sec x+C \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given integral.
Split the numerator into two separate fractions.
Express the fractions using \(\sec x\) and \(\tan x\).
Separate the integral using linearity.
Apply the standard trigonometric integration formulas.
Combine the results, preserving the negative sign of the second term.
Verify the result by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 6 steps

Given —
\[I=\int\frac{2-3\sin x}{\cos^2x}\,dx\]
Split the fraction
Separating the numerator into two terms gives
\[I=\int\left(\frac{2}{\cos^2x}-\frac{3\sin x}{\cos^2x}\right)\,dx\]
Apply trigonometric identities
Since \(\sec x=\frac{1}{\cos x}\),
\[\frac{2}{\cos^2x}=2\sec^2x\]
For the second term, factor the denominator as \(\cos x\cdot\cos x\):
\[ \begin{aligned} \frac{3\sin x}{\cos^2x} &=3\left(\frac{1}{\cos x}\right) \left(\frac{\sin x}{\cos x}\right)\\[5pt] &=3\sec x\tan x \end{aligned} \]
Hence,
\[I=\int(2\sec^2x-3\sec x\tan x)\,dx\]
Separate the integral
Applying the linearity property,
\[I=2\int\sec^2x\,dx-3\int\sec x\tan x\,dx\]
Integrate each term
Using the standard formulas,
\[\int\sec^2x\,dx=\tan x\]
and
\[\int\sec x\tan x\,dx=\sec x,\]
The negative sign before \(3\sec x\) is retained because the original integrand contains the term \(-3\sin x\).

💡 Answer

Final Answer

›

\[\boxed{\int\frac{2-3\sin x}{\cos^2x}\,dx=2\tan x-3\sec x+C}\]

🎯 Exam Significance

Exam Significance

›

Trigonometric identities: Tests the use of \(\sec x=\frac{1}{\cos x}\) and \(\tan x=\frac{\sin x}{\cos x}\).
Algebraic simplification: Demonstrates how splitting a fraction simplifies integration.
Standard formulas: Reinforces the integrals of \(\sec^2x\) and \(\sec x\tan x\).
Sign accuracy: Requires careful handling of the negative coefficient in the second term.
Methodical presentation: Shows the algebraic and integration steps clearly for a well-structured solution.

Significance for Competitive Entrance Examinations

Pattern recognition: Identifying standard trigonometric forms enables quick evaluation.
Efficient simplification: Converting fractions into secant and tangent expressions reduces unnecessary manipulation.
Calculation accuracy: Preserving coefficients and signs prevents common errors in direct integration.
Conceptual fluency: Connects trigonometric identities with differentiation and integration.
Transferable technique: The same approach is useful for many integrals involving powers of \(\cos x\) in the denominator.

🚫 Common Mistakes

Common Mistakes

›

Splitting the fraction incorrectly or failing to divide both numerator terms by \(\cos^2x\).
Writing \(\frac{\sin x}{\cos^2x}=\tan x\) instead of \(\sec x\tan x\).
Using \(\int\sec^2x\,dx=\sec x+C\) instead of \(\tan x+C\).
Changing the negative sign in \(-3\int\sec x\tan x\,dx\).
Omitting the constant of integration \(C\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

Split a rational trigonometric expression into simpler terms whenever possible.
Use reciprocal and quotient identities to express fractions in terms of \(\sec x\) and \(\tan x\).
Remember that \(\int\sec^2x\,dx=\tan x+C\).
Remember that \(\int\sec x\tan x\,dx=\sec x+C\).
Preserve the original signs and numerical coefficients.
Verify the antiderivative by differentiation.

Q21

LONG ANSWER4 marks

The antiderivative of the following function equals which of the given options? \[f(x)=\sqrt{x}+\frac{1}{\sqrt{x}}\]

\[ \frac13x^{1/3}+2x^{1/2}+C \]
\[ \frac23x^{2/3}+\frac12x^2+C \]
\[ \frac23x^{3/2}+2x^{1/2}+C \]
\[ \frac32x^{3/2}+\frac12x^{1/2}+C \]

📘 Concept & Theory

Theory and Concept

›

An antiderivative of a function \(f(x)\) is a function \(F(x)\) whose derivative is \(f(x)\). Finding an antiderivative is the process of indefinite integration.

1. Expressing radicals as fractional powers

\[ \sqrt{x}=x^{1/2}, \qquad \frac{1}{\sqrt{x}}=x^{-1/2} \]

2. Power rule of integration

For \(n\ne -1\),

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

The exponent must be increased by one, and the resulting expression must be divided by the new exponent.

🗺️ Solution Roadmap

Step-by-step Plan

›

Rewrite the square root and reciprocal square root as powers of \(x\).
Separate the integral into two terms.
Apply the power rule to each term.
Simplify the coefficients and combine the results.
Compare the resulting expression with the given options.
Verify the answer by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 5 steps

Given —
\[I=\int\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)\,dx\]
Convert radicals into fractional powers.
Using \(\sqrt{x}=x^{1/2}\) and \(\frac{1}{\sqrt{x}}=x^{-1/2}\), we get
\[I=\int\left(x^{1/2}+x^{-1/2}\right)\,dx\]
Separate the integral
By the linearity property of integration,
\[I=\int x^{1/2}\,dx+\int x^{-1/2}\,dx\]
Integrate the first term
Applying the power rule with \(n=\frac12\),
\[ \begin{aligned} \int x^{1/2}\,dx &=\frac{x^{1/2+1}}{1/2+1}\\[5pt] &=\frac{x^{3/2}}{3/2}\\[5pt] &=\frac23x^{3/2} \end{aligned} \]
Integrate the second term
Applying the power rule with \(n=-\frac12\),
\[ \begin{aligned} \int x^{-1/2}\,dx &=\frac{x^{-1/2+1}}{-1/2+1}\\[5pt] &=\frac{x^{1/2}}{1/2}\\[5pt] &=2x^{1/2} \end{aligned} \]
Notice that the new exponent is \(\frac12\), and division by \(\frac12\) is equivalent to multiplication by \(2\).
Combine the results
Adding both antiderivatives and including the constant of integration \(C\),
\[I=\frac23x^{3/2}+2x^{1/2}+C\]

💡 Answer

Final Answer

›

Correct Option: (C)

\[\boxed{\frac23x^{3/2}+2x^{1/2}+C}\]

🎯 Exam Significance

Exam Significance

›

Power rule: Tests the application of the fundamental formula for integrating powers of \(x\).
Fractional and negative exponents: Reinforces the conversion of radicals and reciprocal radicals into powers.
Multiple-choice questions: Develops accuracy in identifying the correct antiderivative from closely related options.
Verification: Differentiation offers a reliable way to confirm the result.

Significance for Competitive Entrance Examinations

Rapid evaluation: Recognising fractional powers makes direct integration faster.
Exponent accuracy: Tests whether the exponent is increased by one correctly.
Option elimination: Incorrect exponents and coefficients in the other choices can be identified by differentiating them.
Fundamental skill: The power rule is frequently used in more advanced integration problems.

🚫 Common Mistakes

Common Mistakes

›

Writing \(\frac{1}{\sqrt{x}}=x^{1/2}\) instead of \(x^{-1/2}\).
Failing to increase each exponent by one before dividing.
Writing \(\frac{x^{1/2}}{1/2}\) as \(\frac12x^{1/2}\) instead of \(2x^{1/2}\).
Confusing \(x^{3/2}\) with \(x^{2/3}\).
Forgetting the constant of integration \(C\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

Rewrite radicals as fractional powers to simplify integration.
Use the power rule for both positive and negative exponents, provided the exponent is not \(-1\).
Remember that dividing by \(\frac12\) is equivalent to multiplying by \(2\).
Compare the complete expression, including exponents and coefficients, with the answer choices.
Differentiate the proposed antiderivative to verify its correctness.

Q22

LONG ANSWER4 marks

If \(\frac{d}{dx}f(x)=4x^3-\frac{3}{x^4}\;\) such that \(f(2)=0\), then \(f(x)\) is:

\[ x^4+\frac{1}{x^3}-\frac{129}{8} \]
\[ x^3+\frac{1}{x^4}+\frac{129}{8} \]
\[ x^4+\frac{1}{x^3}+\frac{129}{8} \]
\[ x^3+\frac{1}{x^4}-\frac{129}{8} \]

📘 Concept & Theory

Theory and Concept

›

This question tests the relationship between differentiation and integration. If the derivative of a function is known, the original function can be obtained by integration. The given condition \(f(2)=0\) is then used to determine the constant of integration.

1. Integration as the inverse of differentiation

If

\[ f'(x)=g(x), \]

then

\[ f(x)=\int g(x)\,dx. \]

2. Power rule of integration

For \(n\ne -1\),

\[ \int x^n\,dx=\frac{x^{n+1}}{n+1}+C. \]

In particular, the negative power in the given derivative must be handled carefully:

\[ -\frac{3}{x^4}=-3x^{-4}. \]

Therefore,

\[ \begin{aligned} \int -3x^{-4}\,dx &=-3\frac{x^{-3}}{-3}\\[5pt] &=x^{-3}\\[5pt] &=\frac{1}{x^3}. \end{aligned} \]

3. Initial condition

The condition \(f(2)=0\) means that the value of the function at \(x=2\) is zero. Substituting \(x=2\) into the general expression for \(f(x)\) allows us to calculate \(C\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Integrate the given derivative to obtain the general function \(f(x)\).
Apply the power rule to both terms.
Include the constant of integration \(C\).
Use the condition \(f(2)=0\) to determine \(C\).
Substitute the constant into the general function.
Compare the resulting expression with the options and verify by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 7 steps

Given —
\[\frac{d}{dx}f(x)=4x^3-\frac{3}{x^4}\]
Integrate both sides with respect to \(x\).
\[f(x)=\int\left(4x^3-\frac{3}{x^4}\right)\,dx\]
Express the negative power explicitly
Since \(\frac{1}{x^4}=x^{-4}\),
\[ f(x)=\int(4x^3-3x^{-4})\,dx. \]
Applying the power rule,
\[ \begin{aligned} f(x) &=4\frac{x^{3+1}}{3+1} -3\frac{x^{-4+1}}{-4+1}+C\\[5pt] &=4\frac{x^4}{4} -3\frac{x^{-3}}{-3}+C\\[5pt] &=x^4+x^{-3}+C\\[5pt] &=x^4+\frac{1}{x^3}+C. \end{aligned} \]
Thus, the general function is
\[ f(x)=x^4+\frac{1}{x^3}+C. \]
Apply the condition \(f(2)=0\).
Substitute \(x=2\) and \(f(2)=0\) into the general function:
\[ 0=2^4+\frac{1}{2^3}+C. \]
Simplifying,
\[ \begin{aligned} 0&=16+\frac18+C\\[5pt] C&=-16-\frac18\\[5pt] C&=-\frac{128}{8}-\frac18\\[5pt] C&=-\frac{129}{8}. \end{aligned} \]
Determine the particular function
Substitute \(C=-\frac{129}{8}\) into the general function:
\[ f(x)=x^4+\frac{1}{x^3}-\frac{129}{8}. \]

💡 Answer

Final Answer

›

Correct Option: (A)

\[ \boxed{f(x)=x^4+\frac{1}{x^3}-\frac{129}{8}} \]

🎯 Exam Significance

Exam Significance

›

Integration and differentiation: Tests the use of integration to recover a function from its derivative.
Negative exponents: Reinforces correct integration of terms involving reciprocal powers.
Constant of integration: Demonstrates how a given function value determines the constant.
Rational arithmetic: Requires accurate conversion and addition of fractions when calculating \(C\).
Verification: Shows how to confirm both the derivative and the specified function value.

Significance for Competitive Entrance Examinations

Initial-value problems: Develops the ability to determine a unique function from a derivative and a specified value.
Fast option selection: Correctly integrating the derivative and calculating \(C\) identifies the matching choice efficiently.
Exponent manipulation: Tests the power rule with negative exponents.
Accuracy under time pressure: Careful calculation of the integration constant prevents errors in multiple-choice questions.

🚫 Common Mistakes

Common Mistakes

›

Integrating \(-3x^{-4}\) as \(-x^{-3}\) instead of \(+x^{-3}\).
Forgetting the arbitrary constant \(C\) when finding the general function.
Substituting \(x=2\) without using the condition \(f(2)=0\).
Calculating \(16+\frac18\) incorrectly; its value is \(\frac{129}{8}\).
Choosing an option with the wrong sign for the constant.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

Integrate the derivative to recover the general function.
Apply the power rule carefully to negative powers.
Use the given function value to determine the constant of integration.
Verify the answer using both the derivative and the initial condition.
The correct answer is option (A).

Integrals

Standard Integration Formulas

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

Chapter Complete!

Recent posts

INTEGRALS — Learning Resources

Frequently Asked Questions

Let's Connect

Integrals

Standard Integration Formulas

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

Chapter Complete!

Recent posts

INTEGRALS — Learning Resources

Frequently Asked Questions

Q 01 What is covered in NCERT Class 12 Mathematics Chapter 7 Exercise 7.1?

Q 02 How do you solve NCERT Class 12 Integrals Exercise 7.1 questions?

Q 03 What is the power rule of integration?

Q 04 What is the difference between differentiation and integration?

Q 05 Why is the constant of integration necessary?

Q 06 How do you integrate square roots and reciprocal square roots?

Q 07 What are the standard trigonometric integration formulas used in Exercise 7.1?

Q 08 How can you verify an indefinite integral?

Q 09 How do you find a function when its derivative and one function value are given?

Q 10 How is NCERT Class 12 Integrals Exercise 7.1 useful for CBSE and JEE preparation?

Let's Connect