NCERT Class 12 Maths Integrals Exercise 7.4 Solutions

📘 Concept & Theory

Theory and Concept

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This problem illustrates the method of substitution, which is useful when the integrand contains a composite function and its derivative, or a constant multiple of that derivative.

The denominator \(x^6+1\) can be expressed as \((x^3)^2+1\), while the numerator \(3x^2\,dx\) is the differential of \(x^3\). This suggests substituting \(t=x^3\) to transform the given integral into a standard inverse trigonometric integral.

The standard integral formula used here is:

\[ \int \frac{1}{t^2+a^2}\,dt =\frac{1}{a}\tan^{-1}\left(\frac{t}{a}\right)+C, \qquad a>0 \]

For \(a=1\), the formula reduces to:

\[ \int \frac{1}{1+t^2}\,dt=\tan^{-1}t+C \]

Here, \(C\) represents the arbitrary constant of integration.

🗺️ Solution Roadmap

Step-by-step Plan

›

Rewrite \(x^6\) as \((x^3)^2\).
Choose the substitution \(t=x^3\).
Differentiate the substitution to obtain \(dt=3x^2\,dx\).
Replace the numerator and denominator in terms of \(t\).
Apply the standard formula for integrating \(\frac{1}{1+t^2}\).
Substitute \(t=x^3\) back into the result and include the constant of integration.
Verify the result by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 10 steps

Given —
\[I=\int \frac{3x^2}{x^6+1}\,dx\]
Express the denominator in a suitable form.
Since \(x^6=(x^3)^2\), we can write
\[I=\int \frac{3x^2}{(x^3)^2+1}\,dx\]
Apply the substitution
Let
\[t=x^3\]
Differentiating both sides with respect to \(x\), we obtain
\[\frac{dt}{dx}=3x^2\]
Multiplying both sides by \(dx\),
\[dt=3x^2\,dx\]
Thus, the numerator \(3x^2\,dx\) can be replaced by \(dt\), and \(x^3\) can be replaced by \(t\).
Transform the integral
Substituting \(t=x^3\) and \(dt=3x^2\,dx\), we get
\[I=\int \frac{dt}{t^2+1}\]
Equivalently,
\[I=\int \frac{1}{1+t^2}\,dt\]
Use the standard integral formula.
We know that
\[\int \frac{1}{1+t^2}\,dt=\tan^{-1}t+C\]
Therefore,
\[I=\tan^{-1}t+C\]
Substitute the original variable.
Since \(t=x^3\), substituting back gives
\[I=\tan^{-1}(x^3)+C\]
Hence, the required integral is
\[\boxed{\int \frac{3x^2}{x^6+1}\,dx=\tan^{-1}(x^3)+C}\]

🎯 Exam Significance

Exam Significance

›

Substitution method: This problem tests the ability to identify a suitable substitution that simplifies both the numerator and the denominator.
Standard integral formula: Students should remember the integral of \(\frac{1}{1+t^2}\), which frequently appears in integration exercises.
Chain rule connection: Recognising the relationship between differentiation and integration helps students solve composite-function integrals efficiently.
Stepwise presentation: Writing the substitution, its differential, the transformed integral, and the final answer clearly supports a complete mathematical solution.
Constant of integration: The arbitrary constant \(C\) must be included in an indefinite integral.

6. Significance for Competitive Entrance Examinations

JEE Main and Advanced: The problem develops pattern recognition for substitution-based integration and the use of standard inverse trigonometric forms.
NEET and other entrance examinations: Familiarity with standard integrals helps improve accuracy and reduces the time needed for routine integration problems.
Higher-level integration: The same substitution principle is useful when an integral contains a composite expression and a corresponding differential factor.
Derivative-integral relationship: Verification by differentiation is a reliable way to check an antiderivative and identify errors involving powers or coefficients.

🚫 Common Mistakes

Common Mistakes

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Writing \(x^6=x^3+3\) instead of correctly expressing \(x^6=(x^3)^2\).
Forgetting that \(dt=3x^2\,dx\), which would leave an incorrect coefficient in the transformed integral.
Using \(\ln(1+t^2)\) as the antiderivative of \(\frac{1}{1+t^2}\).
Writing \(\tan^{-1}x^3\) ambiguously instead of \(\tan^{-1}(x^3)\).
Omitting the constant of integration \(C\).

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 5 points

The substitution \(t=x^3\) converts the given integral into a standard inverse tangent integral.
The numerator \(3x^2\,dx\) is the differential of \(x^3\), making the substitution direct and convenient.
The key formula is \(\displaystyle \int\frac{1}{1+t^2}\,dt=\tan^{-1}t+C\).
Always substitute the original variable back after integrating with respect to the new variable.
Differentiation provides an effective check of the final answer.

📘 Concept & Theory

Theory and Concept

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This problem illustrates trigonometric substitution and the standard integral involving the reciprocal of the square root of a quadratic expression. The expression \(1+4x^2\) suggests using a substitution involving the tangent function because of the identity

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

By choosing \(2x=\tan\theta\), the expression under the square root becomes a perfect square in terms of \(\sec\theta\). This simplifies the radical and transforms the integral into an integral involving \(\sec\theta\).

The relevant standard integral is

\[ \int \sec\theta\,d\theta =\ln|\sec\theta+\tan\theta|+C \]

Alternatively, the problem can be solved directly using the standard formula

\[ \int\frac{dx}{\sqrt{x^2+a^2}} =\ln\left|x+\sqrt{x^2+a^2}\right|+C, \qquad a>0 \]

In the given integral, the substitution \(u=2x\) reduces it to this standard form. Both methods lead to the same result.

🗺️ Solution Roadmap

Step-by-step Plan

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Identify the quadratic expression \(1+4x^2\) under the square root.
Choose the trigonometric substitution \(2x=\tan\theta\).
Differentiate to express \(dx\) in terms of \(d\theta\).
Use \(1+\tan^2\theta=\sec^2\theta\) to simplify the radical.
Integrate the resulting expression involving \(\sec\theta\).
Express \(\tan\theta\) and \(\sec\theta\) in terms of \(x\).
Write the final result with the constant of integration and verify it by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 13 steps

Given —
\[I=\int\frac{1}{\sqrt{1+4x^2}}\,dx\]
Apply trigonometric substitution
Let
\[2x=\tan\theta\]
Differentiating both sides with respect to \(\theta\), we get
\[2\frac{dx}{d\theta}=\sec^2\theta\]
Therefore,
\[dx=\frac{1}{2}\sec^2\theta\,d\theta\]
Simplify the expression under the square root.
Since \(2x=\tan\theta\), we have
\[4x^2=\tan^2\theta\]
Consequently,
\[\sqrt{1+4x^2}=\sqrt{1+\tan^2\theta}\]
Using the identity
\[1+\tan^2\theta=\sec^2\theta,\]
we obtain
\[\sqrt{1+4x^2}=\sqrt{\sec^2\theta}=|\sec\theta|\]
For the usual trigonometric substitution, we may take \(\theta\) in the interval \((-\pi/2,\pi/2)\). In this interval, \(\sec\theta>0\), so
\[\sqrt{1+4x^2}=\sec\theta\]
Substitute into the integral
Replacing \(dx\) and the square root with their expressions in terms of \(\theta\), we get
\[I=\int\frac{\frac12\sec^2\theta\,d\theta}{\sec\theta}\]
Simplifying,
\[I=\frac12\int\sec\theta\,d\theta\]
Integrate using the standard formula.
We use
\[\int\sec\theta\,d\theta=\ln|\sec\theta+\tan\theta|+C\]
Hence,
\[I=\frac12\ln|\sec\theta+\tan\theta|+C\]
Express the result in terms of \(x\).
From the substitution,
\[\tan\theta=2x\]
Also, since \(\sec^2\theta=1+\tan^2\theta\) and \(\sec\theta>0\),
\[ \begin{aligned} \sec\theta &=\sqrt{1+\tan^2\theta}\\[5pt] &=\sqrt{1+4x^2} \end{aligned} \]
Substituting these expressions into the integral gives
\[I=\frac12\ln\left|\sqrt{1+4x^2}+2x\right|+C\]
Since \(\sqrt{1+4x^2}>|2x|\), the quantity \(\sqrt{1+4x^2}+2x\) is always positive for real \(x\). Therefore, the absolute value signs can be omitted.

💡 Answer

Final Answer

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\[ \boxed{ \displaystyle \int\frac{1}{\sqrt{1+4x^2}}\,dx =\frac12\ln\left(2x+\sqrt{1+4x^2}\right)+C } \]

🎯 Exam Significance

Exam Significance

›

Trigonometric substitution: The problem tests the ability to select a substitution that simplifies a radical expression.
Trigonometric identities: The identity \(1+\tan^2\theta=\sec^2\theta\) is central to the solution.
Standard integral formula: Students should remember the integral of \(\frac{1}{\sqrt{1+x^2}}\) and its related forms.
Logical presentation: Showing the substitution, differential, simplification, integration, and back-substitution makes the solution easier to follow and assess.
Verification: Differentiating the final expression confirms the correctness of the antiderivative.

Significance for Competitive Entrance Examinations

JEE Main and Advanced: The question builds fluency in radical integrals, substitution techniques, and the use of standard results.
NEET and other entrance examinations: Recognising standard integral patterns helps students solve routine integration questions efficiently.
Method selection: Comparing trigonometric substitution with direct substitution helps students choose an efficient approach according to the structure of an integral.
Algebraic accuracy: Correct handling of square roots, coefficients, and logarithms is essential for avoiding common errors in objective questions.

🚫 Common Mistakes

Common Mistakes

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Forgetting the factor \(\frac12\) when differentiating \(2x=\tan\theta\).
Writing \(dx=\frac12\sec^2\theta\,d\theta\) incorrectly as \(dx=\frac12\sec^2\theta\,dt\). The differential must be \(d\theta\).
Forgetting that \(\sqrt{\sec^2\theta}=|\sec\theta|\), and that it equals \(\sec\theta\) only when \(\sec\theta>0\).
Omitting the factor \(\frac12\) in the final logarithmic expression.
Using an incorrect formula for \(\int\sec\theta\,d\theta\).
Forgetting the constant of integration \(C\).

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 6 points

The substitution \(2x=\tan\theta\) transforms the radical into a simpler trigonometric expression.
The identity \(1+\tan^2\theta=\sec^2\theta\) is the key simplification.
The standard result is \(\displaystyle \int\sec\theta\,d\theta=\ln|\sec\theta+\tan\theta|+C\).
The factor \(\frac12\) arises from differentiating \(2x=\tan\theta\) and must be retained throughout the solution.
Direct substitution \(u=2x\) is an alternative efficient method.
Always verify an indefinite integral by differentiating the proposed antiderivative when possible.

📘 Concept & Theory

Theory and Concept

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This problem illustrates the method of substitution combined with a standard integral involving the square root of a quadratic expression. The expression \((2-x)^2+1\) suggests substituting \(t=2-x\), which transforms the quadratic expression into \(t^2+1\).

Since the derivative of \(2-x\) is \(-1\), the substitution introduces a negative sign in the integral. Keeping track of this sign is essential for obtaining the correct antiderivative.

The standard integral formula is

\[ \int \frac{dt}{\sqrt{t^2+a^2}} =\ln\left|t+\sqrt{t^2+a^2}\right|+C, \qquad a>0 \]

For \(a=1\), this becomes

\[ \int \frac{dt}{\sqrt{t^2+1}} =\ln\left|t+\sqrt{t^2+1}\right|+C \]

Therefore, the given integral can be evaluated directly after the substitution \(t=2-x\). A trigonometric substitution, \(t=\tan\theta\), provides an alternative derivation of the same result.

🗺️ Solution Roadmap

Step-by-step Plan

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Identify the expression \(2-x\) inside the square root.
Substitute \(t=2-x\) and calculate \(dt=-dx\).
Rewrite the integral in terms of \(t\), carefully retaining the negative sign.
Apply the standard integral formula for \(\frac{1}{\sqrt{t^2+1}}\).
Substitute \(t=2-x\) back into the result.
Verify the final answer by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 11 steps

Given —
\[I=\int \frac{1}{\sqrt{(2-x)^2+1}}\,dx\]
Apply substitution
Let
\[t=2-x\]
Differentiating both sides with respect to \(x\),
\[\frac{dt}{dx}=-1\]
Therefore,
\[dt=-dx\]
or
\[dx=-dt\]
Transform the integral
Since \(t=2-x\), we have
\[(2-x)^2+1=t^2+1\]
Substituting \(t=2-x\) and \(dx=-dt\) into the integral gives
\[ \begin{aligned} I &=\int\frac{1}{\sqrt{t^2+1}}(-dt)\\[6pt] &=-\int\frac{dt}{\sqrt{t^2+1}} \end{aligned} \]
The negative sign appears because the derivative of \(2-x\) is \(-1\).
Use the standard integral formula
We know that
\[\int\frac{dt}{\sqrt{t^2+1}}=\ln\left|t+\sqrt{t^2+1}\right|+C\]
Hence,
\[I=-\ln\left|t+\sqrt{t^2+1}\right|+C\]
Substitute the original expression
Since \(t=2-x\), substituting back gives
\[I=-\ln\left|2-x+\sqrt{(2-x)^2+1}\right|+C\]
Using the logarithmic identity
\[-\ln|A|=\ln\left|\frac{1}{A}\right|,\quad A\ne 0,\]
we can also express the result as
\[I=\ln\left|\frac{1}{2-x+\sqrt{(2-x)^2+1}}\right|+C\]
Notice that
\[\sqrt{(2-x)^2+1}>|2-x|\]
Consequently, \(2-x+\sqrt{(2-x)^2+1}\) is always positive for real \(x\). Therefore, the absolute value signs around this expression can be omitted.

💡 Answer

Final Answer

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

🎯 Exam Significance

Exam Significance

›

Substitution technique: The problem tests the ability to simplify a composite expression by choosing an appropriate variable.
Handling negative signs: The substitution \(t=2-x\) requires careful differentiation because \(dt=-dx\).
Standard integral: Students should remember the formula for integrating the reciprocal of the square root of \(t^2+a^2\).
Stepwise working: Showing the variable substitution, transformed integral, standard formula, and back-substitution makes the reasoning clear.
Verification: Differentiation confirms that the final expression is a correct antiderivative.

Significance for Competitive Entrance Examinations

JEE Main and Advanced: The problem strengthens skills in substitution, standard integrals, and the handling of composite quadratic expressions.
NEET and other entrance examinations: Familiarity with standard integral forms can improve speed and accuracy in objective questions.
Sign management: Recognising the effect of a negative derivative helps avoid sign errors in more complex integrals.
Method selection: Knowing both direct application of a standard formula and trigonometric substitution helps students choose an efficient solution path.

🚫 Common Mistakes

Common Mistakes

›

Writing \(dx=dt\) instead of \(dx=-dt\).
Dropping the negative sign when transforming the integral.
Confusing \(\sqrt{t^2+1}\) with \(t+1\).
Forgetting the absolute value in the general logarithmic integral formula before checking whether the logarithm's argument is positive.
Using \(dt=\sec^2\theta\,dt\) instead of the correct differential \(dt=\sec^2\theta\,d\theta\).
Omitting the arbitrary constant of integration \(C\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

The substitution \(t=2-x\) converts the integrand into the standard form \(\frac{1}{\sqrt{t^2+1}}\).
The negative sign in \(dx=-dt\) must be retained throughout the integration.
The standard result is \(\displaystyle \int\frac{dt}{\sqrt{t^2+1}}=\ln\left|t+\sqrt{t^2+1}\right|+C\).
Trigonometric substitution \(t=\tan\theta\) offers an alternative derivation using the identity \(1+\tan^2\theta=\sec^2\theta\).
Back-substitution expresses the final answer in terms of the original variable \(x\).
Differentiating the result is a reliable way to verify an indefinite integral.

📘 Concept & Theory

Theory and Concept

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This problem is based on the standard integral involving the square root of the difference of two squares. The expression \(9-25x^2\) can be written as \(3^2-(5x)^2\), which matches the standard form \(a^2-u^2\).

The standard integral formula is

\[ \int\frac{du}{\sqrt{a^2-u^2}} =\sin^{-1}\left(\frac{u}{a}\right)+C, \qquad a>0 \]

In this problem, the expression inside the square root contains \(5x\) rather than \(x\). Therefore, we must account for the factor \(5\) by using the substitution \(u=5x\), which gives \(du=5\,dx\) and \(dx=\frac15\,du\).

This factor is essential for obtaining the correct coefficient in the final answer.

🗺️ Solution Roadmap

Step-by-step Plan

›

Rewrite \(9-25x^2\) as \(3^2-(5x)^2\).
Substitute \(u=5x\) to match the standard integral formula.
Differentiate the substitution to obtain \(dx=\frac15\,du\).
Take the constant factor \(\frac15\) outside the integral.
Apply the standard inverse sine integral formula.
Substitute \(u=5x\) back into the result.
Verify the answer by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 12 steps

Given —
\[I=\int\frac{1}{\sqrt{9-25x^2}}\,dx\]
Express the denominator in standard form
Since \(9=3^2\) and \(25x^2=(5x)^2\), we can write
\[I=\int\frac{1}{\sqrt{3^2-(5x)^2}}\,dx\]
Compare this with the standard form
\[\int\frac{du}{\sqrt{a^2-u^2}}\]
We identify \(a=3\) and choose \(u=5x\).
Apply substitution
Let
\[u=5x\]
Differentiating both sides with respect to \(x\),
\[\frac{du}{dx}=5\]
Therefore,
\[du=5\,dx\]
Hence,
\[dx=\frac15\,du\]
Transform the integral
Substituting \(u=5x\) and \(dx=\frac15\,du\), we obtain
\[I=\int\frac{\frac15\,du}{\sqrt{3^2-u^2}}\]
Taking the constant factor outside the integral gives
\[I=\frac15\int\frac{du}{\sqrt{9-u^2}}\]
Apply the standard integral formula.
Recall that
\[\int\frac{du}{\sqrt{a^2-u^2}}=\sin^{-1}\left(\frac{u}{a}\right)+C\]
Here, \(a=3\). Therefore,
\[\int\frac{du}{\sqrt{9-u^2}}=\sin^{-1}\left(\frac{u}{3}\right)+C\]
Consequently,
\[I=\frac15\sin^{-1}\left(\frac{u}{3}\right)+C\]
Substitute the original variable
Since \(u=5x\), substituting back gives
\[I=\frac15\sin^{-1}\left(\frac{5x}{3}\right)+C\]

💡 Answer

Final Answer

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\[ \boxed{ \displaystyle \int\frac{1}{\sqrt{9-25x^2}}\,dx =\frac15\sin^{-1}\left(\frac{5x}{3}\right)+C } \]

🎯 Exam Significance

Exam Significance

›

Recognition of standard forms: The problem tests the ability to identify an integral of the form \(\frac{1}{\sqrt{a^2-u^2}}\).
Correct substitution: Students must recognise that the expression \(5x\) plays the role of \(u\).
Handling constant factors: The factor \(\frac15\) arises from \(dx=\frac15\,du\) and must not be omitted.
Inverse trigonometric functions: The problem reinforces the relationship between standard integrals and inverse trigonometric functions.
Stepwise presentation: Showing the substitution, differential, formula, and back-substitution makes the solution logically complete.

Significance for Competitive Entrance Examinations

JEE Main and Advanced: This problem builds fluency in standard integral recognition and substitution involving linear expressions.
NEET and other entrance examinations: Memorising and applying standard integral formulas helps students solve objective questions efficiently.
Coefficient accuracy: Correctly handling the constant factor introduced by substitution is essential in timed examinations.
Inverse trigonometric integration: The problem provides practice in connecting algebraic radical expressions with inverse sine functions.
Answer verification: Differentiation helps identify errors in coefficients and arguments of inverse trigonometric functions.

🚫 Common Mistakes

Common Mistakes

›

Writing \(dx=du\) instead of \(dx=\frac15\,du\).
Forgetting the coefficient \(\frac15\) in the final answer.
Writing \(\sin^{-1}(5x)\) instead of \(\sin^{-1}\left(\frac{5x}{3}\right)\).
Confusing the formula for \(\frac{1}{\sqrt{a^2-u^2}}\) with the formula for \(\frac{1}{a^2+u^2}\).
Omitting the constant of integration \(C\).
Ignoring the real-domain condition \(9-25x^2>0\) when considering the integrand as a real-valued function.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

The integral matches the standard form \(\displaystyle\int\frac{du}{\sqrt{a^2-u^2}}\).
The substitution \(u=5x\) simplifies the expression under the square root.
The differential relation \(dx=\frac15\,du\) introduces the essential coefficient \(\frac15\).
The standard result is \(\displaystyle\int\frac{du}{\sqrt{a^2-u^2}}=\sin^{-1}\left(\frac{u}{a}\right)+C\).
The final answer must contain the argument \(\frac{5x}{3}\) inside the inverse sine function.
Differentiation provides a direct check of the obtained antiderivative.

📘 Concept & Theory

Theory and Concept

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This problem is based on the method of substitution and the standard integral involving the inverse tangent function. The denominator \(1+2x^4\) can be expressed as

\[ 1+2x^4=1+\left(\sqrt{2}x^2\right)^2 \]

This suggests the substitution \(t=\sqrt{2}x^2\). Differentiating this expression gives a differential involving \(x\,dx\), which appears in the numerator of the original integral. Thus, the substitution transforms the integral into the standard form

\[ \int\frac{dt}{1+t^2}=\tan^{-1}t+C \]

The key is to choose a substitution that simplifies the denominator while simultaneously converting the numerator and differential into a constant multiple of \(dt\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Rewrite \(2x^4\) as \((\sqrt{2}x^2)^2\).
Choose the substitution \(t=\sqrt{2}x^2\).
Differentiate to express \(x\,dx\) in terms of \(dt\).
Substitute into the integral and simplify the constant coefficient.
Apply the standard inverse tangent integral formula.
Substitute \(t=\sqrt{2}x^2\) back into the result.
Verify the answer by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 12 steps

Given —
\[I=\int\frac{3x}{1+2x^4}\,dx\]
Express the denominator in a suitable form
Since
\[2x^4=(\sqrt{2}x^2)^2,\]
we can rewrite the integral as
\[I=\int\frac{3x}{1+(\sqrt{2}x^2)^2}\,dx\]
Apply substitution
Let
\[t=\sqrt{2}x^2\]
Differentiating both sides with respect to \(x\),
\[\frac{dt}{dx}=2\sqrt{2}x\]
Therefore,
\[dt=2\sqrt{2}x\,dx\]
Rearranging,
\[x\,dx=\frac{dt}{2\sqrt{2}}\]
Transform the integral
The original integral is
\[I=\int\frac{3x}{1+(\sqrt{2}x^2)^2}\,dx\]
Substituting \(t=\sqrt{2}x^2\) and \(x\,dx=\frac{dt}{2\sqrt{2}}\), we obtain
\[I=\int\frac{3}{1+t^2}\cdot\frac{dt}{2\sqrt{2}}\]
Taking the constant factor outside the integral,
\[I=\frac{3}{2\sqrt{2}}\int\frac{dt}{1+t^2}\]
Apply the standard integral formula.
We know that
\[\int\frac{dt}{1+t^2}=\tan^{-1}t+C\]
Therefore,
\[I=\frac{3}{2\sqrt{2}}\tan^{-1}t+C\]
Substitute the original variable.
Since \(t=\sqrt{2}x^2\), substituting back gives
\[I=\frac{3}{2\sqrt{2}}\tan^{-1}(\sqrt{2}x^2)+C\]

💡 Answer

Final Answer

›

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

🎯 Exam Significance

Exam Significance

›

Substitution method: This question tests the ability to identify a substitution that simplifies a polynomial expression in the denominator.
Standard integral recognition: Students must recognise the form \(\frac{1}{1+t^2}\), whose antiderivative is an inverse tangent function.
Correct differential handling: Deriving \(x\,dx=\frac{dt}{2\sqrt{2}}\) is an essential step in obtaining the correct coefficient.
Stepwise presentation: Showing the substitution, differential, transformed integral, and back-substitution makes the solution clear and logically complete.
Verification: Differentiation confirms the correctness of the antiderivative and reinforces the relationship between differentiation and integration.

Significance for Competitive Entrance Examinations

JEE Main and Advanced: The problem develops pattern recognition for integrals that can be reduced to standard inverse trigonometric forms.
NEET and other entrance examinations: Recognising standard integral forms helps students solve routine objective questions accurately and efficiently.
Algebraic manipulation: Rewriting \(2x^4\) as \((\sqrt{2}x^2)^2\) illustrates how a suitable algebraic transformation can simplify integration.
Coefficient accuracy: Correctly handling numerical factors during substitution is important for avoiding errors in multiple-choice questions.
Transferable technique: The same approach applies to other integrals in which the denominator has the form \(1+[f(x)]^2\) and the numerator contains a constant multiple of \(f'(x)\).

🚫 Common Mistakes

Common Mistakes

›

Writing \(2x^4=(\sqrt{2}x^2)\) instead of \(2x^4=(\sqrt{2}x^2)^2\).
Forgetting the factor \(2\sqrt{2}\) when differentiating \(t=\sqrt{2}x^2\).
Writing \(x\,dx=\frac{dt}{\sqrt{2}}\) instead of the correct expression \(x\,dx=\frac{dt}{2\sqrt{2}}\).
Omitting the coefficient \(\frac{3}{2\sqrt{2}}\) in the final answer.
Using the formula for \(\int\frac{dt}{1+t^2}\) incorrectly.
Forgetting the constant of integration \(C\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

The substitution \(t=\sqrt{2}x^2\) transforms the denominator into \(1+t^2\).
The differential relation \(x\,dx=\frac{dt}{2\sqrt{2}}\) determines the constant coefficient.
The standard result is \(\displaystyle\int\frac{dt}{1+t^2}=\tan^{-1}t+C\).
The inverse tangent function appears because the transformed integrand has the standard form \(\frac{1}{1+t^2}\).
Substitute the original variable back into the antiderivative before presenting the final answer.
Differentiate the final expression to verify the result whenever possible.

📘 Concept & Theory

Theory and Concept

›

This problem is based on the method of substitution and the standard integral involving the difference of two squares. The denominator \(1-x^6\) can be rewritten as \(1-(x^3)^2\). Since the numerator contains \(x^2\), the substitution \(t=x^3\) is appropriate because its derivative contains the factor \(3x^2\).

After substitution, the integral reduces to the standard form

\[ \int\frac{dt}{a^2-t^2} =\frac{1}{2a}\ln\left|\frac{a+t}{a-t}\right|+C \]

An equivalent form is

\[ \int\frac{dt}{a^2-t^2} =\frac{1}{2a}\ln\left|\frac{a+t}{a-t}\right|+C \]

For \(a=1\), we obtain

\[ \int\frac{dt}{1-t^2} =\frac12\ln\left|\frac{1+t}{1-t}\right|+C \]

The logarithmic ratio may also be written with the numerator and denominator interchanged, provided the overall sign is reversed. This sign is important when simplifying the final result.

🗺️ Solution Roadmap

Step-by-step Plan

›

Rewrite \(x^6\) as \((x^3)^2\).
Choose the substitution \(t=x^3\).
Differentiate to obtain \(x^2\,dx=\frac13\,dt\).
Transform the integral into \(\frac13\int\frac{dt}{1-t^2}\).
Apply the standard logarithmic integral formula.
Substitute \(t=x^3\) back into the result.
Verify the final expression by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 11 steps

Given —
\[I=\int\frac{x^2}{1-x^6}\,dx\]
Express the denominator in a suitable form
Since \(x^6=(x^3)^2\), we can write
\[I=\int\frac{x^2}{1-(x^3)^2}\,dx\]
The expression \(x^3\) appears inside the denominator, and its derivative is closely related to the numerator. Therefore, we use substitution.
Apply substitution
Let
\[t=x^3\]
Differentiating both sides with respect to \(x\),
\[\frac{dt}{dx}=3x^2\]
Therefore,
\[dt=3x^2\,dx\]
Rearranging,
\[x^2\,dx=\frac{dt}{3}\]
Transform the integral
Substituting \(t=x^3\) and \(x^2\,dx=\frac13\,dt\), we obtain
\[I=\int\frac{\frac13\,dt}{1-t^2}\]
Taking the constant factor outside the integral,
\[I=\frac13\int\frac{dt}{1-t^2}\]
Apply the standard integral formula
The standard result is
\[\int\frac{dt}{a^2-t^2}=\frac{1}{2a}\ln\left|\frac{a+t}{a-t}\right|+C\]
Taking \(a=1\), we obtain
\[\int\frac{dt}{1-t^2}=\frac12\ln\left|\frac{1+t}{1-t}\right|+C\]
Therefore,
\[I=\frac16\ln\left|\frac{1+t}{1-t}\right|+C\]
Substitute the original variable.
Since \(t=x^3\), substituting back gives
\[I=\frac16\ln\left|\frac{1+x^3}{1-x^3}\right|+C\]
This is equivalent to writing the logarithmic ratio in reciprocal form with a negative sign:
\[I=-\frac16\ln\left|\frac{1-x^3}{1+x^3}\right|+C\]

💡 Answer

Final Answer

›

\[ \boxed{ \displaystyle \int\frac{x^2}{1-x^6}\,dx =\frac16\ln\left|\frac{1+x^3}{1-x^3}\right|+C } \]

🎯 Exam Significance

Exam Significance

›

Substitution method: This problem tests the ability to identify a substitution that simplifies a polynomial expression in the denominator.
Standard logarithmic integral: Students should recognise the form \(\frac{1}{a^2-t^2}\) and recall its logarithmic antiderivative.
Coefficient accuracy: The factor \(\frac13\) comes from the differential relation and must be retained to obtain the correct answer.
Sign accuracy: Interchanging the numerator and denominator inside a logarithm changes the sign. Careful handling of this sign is essential.
Stepwise presentation: Clearly writing the substitution, transformed integral, formula, and final expression demonstrates the complete mathematical method.

Significance for Competitive Entrance Examinations

JEE Main and Advanced: The question develops fluency in substitution and the recognition of rational expressions that lead to logarithmic antiderivatives.
NEET and other entrance examinations: Familiarity with standard integrals improves speed and accuracy in objective questions.
Algebraic simplification: Rewriting \(1-x^6\) as \(1-(x^3)^2\) reveals the structure required for the standard formula.
Sign management: The problem reinforces the importance of handling logarithmic ratios correctly, a common source of errors in integration.
Verification skills: Differentiating the final expression helps students check their answers independently.

🚫 Common Mistakes

Common Mistakes

›

Forgetting the factor \(3\) when differentiating \(t=x^3\).
Writing \(x^2\,dx=3\,dt\) instead of \(x^2\,dx=\frac13\,dt\).
Using the incorrect formula for \(\int\frac{dt}{1-t^2}\).
Interchanging \(1+t\) and \(1-t\) in the logarithmic ratio without changing the overall sign.
Omitting absolute value signs from the general logarithmic result when the sign of the argument has not been established.
Forgetting the constant of integration \(C\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

The substitution \(t=x^3\) is effective because its differential contains the numerator factor \(x^2\,dx\).
The integral reduces to the standard form \(\displaystyle\int\frac{dt}{1-t^2}\).
The standard result is \(\displaystyle\int\frac{dt}{1-t^2}=\frac12\ln\left|\frac{1+t}{1-t}\right|+C\).
The factor \(\frac13\) from substitution combines with the factor \(\frac12\) from the standard formula to give \(\frac16\).
Changing the order of the numerator and denominator inside the logarithm requires reversing the sign.
Differentiation confirms that the obtained antiderivative is correct.

📘 Concept & Theory

Theory and Concept

›

This problem illustrates the method of splitting an integral into simpler parts, followed by substitution and the application of a standard integral formula.

First, split the numerator \(x-1\) into two terms:

\[ I=\int\frac{x}{\sqrt{x^2-1}}\,dx -\int\frac{1}{\sqrt{x^2-1}}\,dx \]

The first integral can be evaluated using the substitution \(t=x^2-1\), because the derivative of \(x^2-1\) is \(2x\), and the numerator contains \(x\).

The second integral is a standard logarithmic integral:

\[ \int\frac{dx}{\sqrt{x^2-a^2}} =\ln\left|x+\sqrt{x^2-a^2}\right|+C \]

For \(a=1\), this becomes

\[ \int\frac{dx}{\sqrt{x^2-1}} =\ln\left|x+\sqrt{x^2-1}\right|+C \]

Important correction: The supplied working evaluates only the first integral after splitting the numerator. The second integral must also be evaluated and subtracted to obtain the complete answer.

🗺️ Solution Roadmap

Step-by-step Plan

›

Split the numerator \(x-1\) into \(x\) and \(-1\).
Separate the original integral into two simpler integrals.
Evaluate the first integral using \(t=x^2-1\).
Evaluate the second integral using the standard logarithmic formula.
Combine both results, keeping track of the negative sign.
Verify the final answer by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 16 steps

Given —
\[I=\int\frac{x-1}{\sqrt{x^2-1}}\,dx\]
Split the numerator
Using the linearity of integration,
\[I=\int\frac{x}{\sqrt{x^2-1}}\,dx-\int\frac{1}{\sqrt{x^2-1}}\,dx\]
Let
\[I=I_1-I_2\]
where
\[I_1=\int\frac{x}{\sqrt{x^2-1}}\,dx\]
and
\[I_2=\int\frac{1}{\sqrt{x^2-1}}\,dx\]
Part A: Evaluate the First Integral
Apply substitution
Consider
\[I_1=\int\frac{x}{\sqrt{x^2-1}}\,dx\]
Let
\[t=x^2-1\]
Differentiating both sides with respect to \(x\),
\[\frac{dt}{dx}=2x\]
Therefore,
\[dt=2x\,dx\]
Hence,
\[x\,dx=\frac12\,dt\]
Transform the integral
Substituting \(t=x^2-1\) and \(x\,dx=\frac12\,dt\), we obtain
\[I_1=\frac12\int\frac{dt}{\sqrt{t}}\]
Expressing the square root as a fractional power,
\[I_1=\frac12\int t^{-1/2}\,dt\]
Apply the power rule of integration.
The standard formula is
\[\int t^n\,dt=\frac{t^{n+1}}{n+1}+C,\quad n\ne-1\]
Taking \(n=-\frac12\), we get
\[ \begin{aligned} I_1 &=\frac12\cdot \frac{t^{-1/2+1}}{-1/2+1}\\[6pt] &=\frac12\cdot\frac{t^{1/2}}{1/2}\\[6pt] &=\frac12\cdot2t^{1/2}\\[6pt] &=\sqrt{t} \end{aligned} \]
Substitute back
Since \(t=x^2-1\),
\[I_1=\sqrt{x^2-1}\]
Part B: Evaluate the Second Integral
Apply the standard logarithmic formula.
Consider
\[I_2=\int\frac{1}{\sqrt{x^2-1}}\,dx\]
Using the standard integral formula
\[\int\frac{dx}{\sqrt{x^2-a^2}}=\ln\left|x+\sqrt{x^2-a^2}\right|+C,\]
and taking \(a=1\), we obtain
\[I_2=\ln\left|x+\sqrt{x^2-1}\right|\]
Part C: Combine Both Results
Substitute \(I_1\) and \(I_2\) into the original integral.
Since \(I=I_1-I_2\),
\[I=\sqrt{x^2-1}-\ln\left|x+\sqrt{x^2-1}\right|+C\]

💡 Answer

Final Answer

›

\[ \boxed{ \displaystyle \int\frac{x-1}{\sqrt{x^2-1}}\,dx =\sqrt{x^2-1} -\ln\left|x+\sqrt{x^2-1}\right|+C } \]

🎯 Exam Significance

Exam Significance

›

Linearity of integration: The problem demonstrates how splitting a numerator into simpler terms can make an integral easier to evaluate.
Substitution technique: The first integral is a direct application of substitution followed by the power rule.
Standard logarithmic integral: The second integral tests familiarity with the standard form involving \(\sqrt{x^2-a^2}\).
Completeness of working: Both integrals must be evaluated. Solving only the first part gives an incomplete answer.
Sign accuracy: The second integral is subtracted, so its logarithmic contribution must retain the negative sign.

Significance for Competitive Entrance Examinations

JEE Main and Advanced: The problem develops the ability to combine substitution, power-rule integration, and standard logarithmic results.
NEET and other entrance examinations: Recognising standard integral forms helps students solve integration questions efficiently.
Strategic decomposition: Splitting a complicated numerator into simpler terms is a useful technique for many rational and radical integrals.
Error prevention: Evaluating each part separately reduces the likelihood of overlooking a term or reversing a sign.
Derivative-integral connection: Verification strengthens accuracy and provides a way to check a proposed antiderivative.

🚫 Common Mistakes

Common Mistakes

›

Evaluating only \(\int\frac{x}{\sqrt{x^2-1}}\,dx\) and forgetting the second integral.
Forgetting the factor \(\frac12\) when using \(t=x^2-1\).
Applying the power rule incorrectly to \(t^{-1/2}\).
Omitting the negative sign before the logarithmic term.
Confusing the formula for \(\int\frac{dx}{\sqrt{x^2-a^2}}\) with the formula for \(\int\frac{dx}{\sqrt{x^2+a^2}}\).
Forgetting the constant of integration \(C\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

Split the numerator \(x-1\) to separate the integral into two manageable parts.
Use \(t=x^2-1\) to evaluate the first integral.
Use the standard logarithmic formula to evaluate the second integral.
Retain the negative sign when combining the two antiderivatives.
Always evaluate every term produced by splitting an integral.
Differentiate the final result to verify that it reproduces the original integrand.

📘 Concept & Theory

Theory and Concept

›

This problem illustrates the method of substitution followed by the application of a standard integral involving a quadratic denominator. The denominator \(x^6+a^6\) can be expressed as the sum of two squares:

\[ x^6+a^6=(x^3)^2+(a^3)^2 \]

Since the numerator contains \(x^2\), the substitution \(t=x^3\) is appropriate. Differentiating this substitution gives \(dt=3x^2\,dx\), allowing the numerator and differential to be replaced by a constant multiple of \(dt\).

The transformed integral is evaluated using the standard formula

\[ \int\frac{dt}{t^2+b^2} =\frac{1}{b}\tan^{-1}\left(\frac{t}{b}\right)+C, \qquad b>0 \]

Here, \(b=a^3\) when \(a>0\). Therefore, the denominator requires the inverse tangent formula, not the logarithmic formula for a denominator of the form \(t^2-b^2\).

Important correction: The supplied working changes the denominator to \(t^2+1\) without dividing by \(a^6\), and later uses a logarithmic expression. These steps are incorrect for the given integral. The correct solution must retain the parameter \(a\) and use the inverse tangent formula.

🗺️ Solution Roadmap

Step-by-step Plan

›

Rewrite \(x^6+a^6\) as \((x^3)^2+(a^3)^2\).
Substitute \(t=x^3\).
Use \(x^2\,dx=\frac13\,dt\) to transform the numerator and differential.
Obtain the integral \(\frac13\int\frac{dt}{t^2+a^6}\).
Apply the standard inverse tangent integral formula with \(b=a^3\), assuming \(a>0\).
Substitute \(t=x^3\) back into the result.
Verify the final expression by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 11 steps

Given —
\[I=\int\frac{x^2}{x^6+a^6}\,dx\]
Express the denominator in a suitable form
Since \(x^6=(x^3)^2\) and \(a^6=(a^3)^2\), we can write
\[I=\int\frac{x^2}{(x^3)^2+(a^3)^2}\,dx\]
Apply substitution
Let
\[t=x^3\]
Differentiating both sides with respect to \(x\),
\[\frac{dt}{dx}=3x^2\]
Therefore,
\[dt=3x^2\,dx\]
Rearranging,
\[x^2\,dx=\frac13\,dt\]
Transform the integral
Substituting \(t=x^3\) and \(x^2\,dx=\frac13\,dt\), we obtain
\[I=\frac13\int\frac{dt}{t^2+a^6}\]
Since \(a^6=(a^3)^2\), the integral becomes
\[I=\frac13\int\frac{dt}{t^2+(a^3)^2}\]
Apply the standard integral formula.
We use
\[\int\frac{dt}{t^2+b^2}=\frac1b\tan^{-1}\left(\frac{t}{b}\right)+C,\qquad b>0\]
Taking \(b=a^3\), assuming \(a>0\), gives
\[\int\frac{dt}{t^2+a^6}=\frac{1}{a^3}\tan^{-1}\left(\frac{t}{a^3}\right)+C\]
Therefore,
\[I=\frac{1}{3a^3}\tan^{-1}\left(\frac{t}{a^3}\right)+C\]
Substitute the original variable
Since \(t=x^3\), substituting back gives
\[I=\frac{1}{3a^3}\tan^{-1}\left(\frac{x^3}{a^3}\right)+C\]

💡 Answer

Final Answer

›

\[ \boxed{ \displaystyle \int\frac{x^2}{x^6+a^6}\,dx =\frac{1}{3a^3}\tan^{-1}\left(\frac{x^3}{a^3}\right)+C, \quad a>0 } \]

🎯 Exam Significance

Exam Significance

›

Substitution method: The question tests the ability to identify a substitution that simplifies a sixth-degree polynomial expression.
Standard integral recognition: Students must identify the form \(\frac{1}{t^2+b^2}\) and apply the inverse tangent formula.
Parameter handling: Retaining \(a^6=(a^3)^2\) is essential. Dropping the parameter leads to an incorrect answer.
Coefficient accuracy: The factor \(\frac13\) from substitution combines with \(\frac{1}{a^3}\) from the standard formula.
Stepwise presentation: Showing the substitution, transformed integral, formula, and back-substitution makes the solution logically complete.

Significance for Competitive Entrance Examinations

JEE Main and Advanced: This problem develops the ability to reduce higher-degree polynomial integrals to standard forms.
NEET and other entrance examinations: Recognising the inverse tangent pattern helps students solve standard integration questions efficiently.
Algebraic manipulation: Expressing the denominator as a sum of squares makes the appropriate standard formula easier to identify.
Parameter awareness: Questions containing a parameter test whether students can preserve constants and coefficients throughout a calculation.
Error detection: Differentiation is an effective way to identify incorrect logarithmic expressions or missing powers of a parameter.

🚫 Common Mistakes

Common Mistakes

›

Replacing \(t^2+a^6\) with \(t^2+1\) without a valid scaling substitution.
Using a logarithmic formula for \(\frac{1}{t^2+a^6}\). This denominator requires an inverse tangent formula.
Forgetting the factor \(\frac13\) when substituting \(t=x^3\).
Omitting the factor \(\frac{1}{a^3}\) in the standard integral formula.
Writing \(\tan^{-1}(x^3)\) instead of the correct argument \(\tan^{-1}\left(\frac{x^3}{a^3}\right)\).
Forgetting the constant of integration \(C\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

The substitution \(t=x^3\) is effective because the numerator contains \(x^2\,dx\).
The transformed integral is \(\displaystyle\frac13\int\frac{dt}{t^2+a^6}\).
The standard formula is \(\displaystyle\int\frac{dt}{t^2+b^2}=\frac1b\tan^{-1}\left(\frac{t}{b}\right)+C\).
The parameter \(a\) must be retained throughout the solution.
The correct antiderivative involves the inverse tangent function, not a logarithm.
Differentiation confirms the final result.

📘 Concept & Theory

Theory and Concept

›

This problem is based on the method of substitution and a standard integral involving the reciprocal of the square root of a quadratic expression.

The integrand contains \(\tan x\) inside the square root and \(\sec^2 x\) as a factor in the numerator. Since

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

the natural substitution is \(t=\tan x\). This transforms the trigonometric integral into an algebraic integral involving \(t\).

The standard formula required is

\[ \int\frac{dt}{\sqrt{t^2+a^2}} =\ln\left|t+\sqrt{t^2+a^2}\right|+C, \qquad a>0 \]

Here, \(a=2\). Therefore, the integral can be evaluated directly using the standard logarithmic formula.

Important correction: The transformed integral is not \(\int\frac{dt}{t^2+4}\). The square root in the denominator must be retained, so the correct form is \(\int\frac{dt}{\sqrt{t^2+4}}\). This changes the applicable standard formula.

🗺️ Solution Roadmap

Step-by-step Plan

›

Identify the relationship between \(\tan x\) and its derivative \(\sec^2 x\).
Substitute \(t=\tan x\).
Use \(dt=\sec^2 x\,dx\) to transform the integral.
Apply the standard logarithmic integral formula for \(\frac{1}{\sqrt{t^2+a^2}}\).
Substitute \(t=\tan x\) back into the result.
Verify the final answer by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 10 steps

Given —
\[I=\int\frac{\sec^2 x}{\sqrt{\tan^2 x+4}}\,dx\]
Apply substitution
Let
\[t=\tan x\]
Differentiating both sides with respect to \(x\),
\[\frac{dt}{dx}=\sec^2 x\]
Therefore,
\[dt=\sec^2 x\,dx\]
Transform the integral
Substituting \(t=\tan x\) and \(dt=\sec^2 x\,dx\), we obtain
\[I=\int\frac{dt}{\sqrt{t^2+4}}\]
Since \(4=2^2\), this becomes
\[I=\int\frac{dt}{\sqrt{t^2+2^2}}\]
Apply the standard integral formula
Recall that
\[\int\frac{dt}{\sqrt{t^2+a^2}}=\ln\left|t+\sqrt{t^2+a^2}\right|+C\]
Taking \(a=2\), we obtain
\[I=\ln\left|t+\sqrt{t^2+4}\right|+C\]
Substitute the original variable
Since \(t=\tan x\), substituting back gives
\[I=\ln\left|\tan x+\sqrt{\tan^2 x+4}\right|+C\]
The expression inside the logarithm is always positive because
\[\sqrt{\tan^2 x+4}>|\tan x|\]
Therefore,
\[\tan x+\sqrt{\tan^2 x+4}>0\]
Hence, the absolute value signs can be omitted wherever the original integrand is defined.

💡 Answer

Final Answer

›

\[ \boxed{ \displaystyle \int\frac{\sec^2 x}{\sqrt{\tan^2 x+4}}\,dx =\ln\left(\tan x+\sqrt{\tan^2 x+4}\right)+C } \]

🎯 Exam Significance

Exam Significance

›

Substitution technique: The question tests recognition of a function and its derivative appearing together in an integrand.
Standard integral formula: Students must distinguish between integrals involving \(\frac{1}{t^2+a^2}\) and \(\frac{1}{\sqrt{t^2+a^2}}\).
Trigonometric differentiation: The identity \(\frac{d}{dx}(\tan x)=\sec^2 x\) is essential to the substitution.
Logical presentation: Showing the substitution, transformed integral, standard formula, and back-substitution provides a complete solution.
Verification: Differentiating the final expression checks the correctness of the logarithmic antiderivative.

Significance for Competitive Entrance Examinations

JEE Main and Advanced: The problem develops fluency in trigonometric substitution and standard radical integrals.
NEET and other entrance examinations: Recognising a function and its derivative helps students solve integration questions efficiently.
Formula selection: Identifying the square root in the denominator prevents the incorrect use of the inverse tangent formula.
Time management: Direct substitution reduces a trigonometric integral to a familiar algebraic form.
Error prevention: Checking the derivative helps students detect mistakes involving the radical or the logarithmic expression.

🚫 Common Mistakes

Common Mistakes

›

Writing \(dt=\sec x\,dx\) instead of \(dt=\sec^2 x\,dx\).
Removing the square root from the denominator during substitution.
Using \(\tan^{-1}t\) as the antiderivative of \(\frac{1}{\sqrt{t^2+4}}\).
Using the logarithmic formula for \(\frac{1}{t^2+4}\) instead of the correct formula for \(\frac{1}{\sqrt{t^2+4}}\).
Forgetting to substitute \(t=\tan x\) back into the final expression.
Omitting the constant of integration \(C\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

The substitution \(t=\tan x\) is appropriate because its derivative is \(\sec^2 x\).
The transformed integral is \(\displaystyle\int\frac{dt}{\sqrt{t^2+4}}\).
The standard formula produces a logarithmic antiderivative, not an inverse tangent function.
The correct result is \(\displaystyle\ln\left|\tan x+\sqrt{\tan^2 x+4}\right|+C\).
Always distinguish between a quadratic denominator and the square root of a quadratic expression.
Differentiation provides a reliable check of the final result.

📘 Concept & Theory

Theory and Concept

›

This problem is based on completing the square and applying a standard integral formula involving the square root of a quadratic expression.

The quadratic expression \(x^2+2x+2\) can be rewritten as a perfect square plus a positive constant:

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

This transformation brings the integral into the standard form

\[ \int\frac{du}{\sqrt{u^2+a^2}} =\ln\left|u+\sqrt{u^2+a^2}\right|+C, \qquad a>0 \]

The appropriate substitution is \(u=x+1\), for which \(du=dx\). The integral can then be evaluated directly using the standard formula.

Important correction: The given working incorrectly identifies the integral as one involving \(x^2+a^2\) without a square root. The denominator contains a square root, so the correct formula is the one involving \(\frac{1}{\sqrt{u^2+a^2}}\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given integral.
Complete the square in \(x^2+2x+2\).
Rewrite the integral using \((x+1)^2+1\).
Substitute \(u=x+1\) and note that \(du=dx\).
Apply the standard logarithmic integral formula.
Substitute \(u=x+1\) back into the result.
Verify the answer by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 14 steps

Given —
\[I=\int\frac{1}{\sqrt{x^2+2x+2}}\,dx\]
Complete the square
Consider the quadratic expression
\[x^2+2x+2\]
We know that
\[x^2+2x+2=x^2+2x+1+1\]
Since
\[x^2+2x+1=(x+1)^2,\]
we obtain
\[x^2+2x+2=(x+1)^2+1\]
Therefore, the integral becomes
\[I=\int\frac{dx}{\sqrt{(x+1)^2+1}}\]
Apply substitution
Let
\[u=x+1\]
Differentiating both sides,
\[du=dx\]
Substituting \(u=x+1\) and \(du=dx\), we get
\[I=\int\frac{du}{\sqrt{u^2+1}}\]
Apply the standard integral formula.
We use the standard result
\[\int\frac{du}{\sqrt{u^2+a^2}}=\ln\left|u+\sqrt{u^2+a^2}\right|+C\]
Here, \(a=1\). Hence,
\[I=\ln\left|u+\sqrt{u^2+1}\right|+C\]
Substitute the original variable.
Since \(u=x+1\), we obtain
\[I=\ln\left|x+1+\sqrt{(x+1)^2+1}\right|+C\]
Simplifying the expression under the square root,
\[(x+1)^2+1=x^2+2x+2\]
Therefore,
\[I=\ln\left|x+1+\sqrt{x^2+2x+2}\right|+C\]
Since
\[\sqrt{(x+1)^2+1}>|x+1|,\]
the expression \(x+1+\sqrt{(x+1)^2+1}\) is always positive. Hence, the absolute value signs can be omitted.

💡 Answer

Final Answer

›

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

🎯 Exam Significance

Exam Significance

›

Completing the square: This problem tests the ability to transform a quadratic expression into a standard form.
Standard integral recognition: Students must identify the formula involving \(\frac{1}{\sqrt{u^2+a^2}}\).
Substitution: The change of variable \(u=x+1\) simplifies the integral without introducing any additional coefficient.
Formula accuracy: Distinguishing between \(\int\frac{du}{u^2+a^2}\) and \(\int\frac{du}{\sqrt{u^2+a^2}}\) is essential.
Stepwise presentation: Showing the square completion, substitution, formula, and back-substitution makes the solution complete and easy to follow.

Significance for Competitive Entrance Examinations

JEE Main and Advanced: The problem strengthens skills in algebraic transformation and the recognition of standard radical integrals.
NEET and other entrance examinations: Familiarity with standard integral forms helps students solve routine integration questions efficiently.
Algebraic simplification: Completing the square is a useful technique for many integrals involving quadratic polynomials.
Formula selection: Correctly identifying the square root in the denominator prevents the use of an inappropriate integral formula.
Verification: Differentiating the final expression helps detect algebraic and calculus errors.

🚫 Common Mistakes

Common Mistakes

›

Writing \(x^2+2x+2=(x+1)^2\), forgetting the additional \(+1\).
Applying the formula for \(\int\frac{dx}{x^2+a^2}\) instead of the formula for \(\int\frac{dx}{\sqrt{x^2+a^2}}\).
Forgetting that \(du=dx\) when \(u=x+1\).
Leaving the final expression in terms of \(u\) rather than returning to \(x\).
Introducing an unnecessary coefficient into the final logarithmic expression.
Omitting the constant of integration \(C\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

Complete the square to rewrite \(x^2+2x+2\) as \((x+1)^2+1\).
The substitution \(u=x+1\) transforms the integral into a standard form.
The applicable formula is \(\displaystyle\int\frac{du}{\sqrt{u^2+a^2}}=\ln\left|u+\sqrt{u^2+a^2}\right|+C\).
The square root in the denominator determines which standard integral formula must be used.
The logarithm's argument is positive for all real \(x\), so the absolute value signs can be omitted.
Differentiation confirms the correctness of the final antiderivative.

📘 Concept & Theory

Theory and Concept

›

This problem is based on completing the square and applying the standard integral formula for an inverse tangent function. The denominator is a quadratic polynomial that can be rewritten as the sum of a perfect square and a positive constant.

The relevant standard formula is

\[ \int\frac{du}{u^2+a^2} =\frac{1}{a}\tan^{-1}\left(\frac{u}{a}\right)+C, \qquad a>0 \]

To use this formula, we first complete the square in \(9x^2+6x+5\). Factoring out \(9\) from the quadratic terms and rewriting the expression produces a denominator involving \(\left(x+\frac13\right)^2+\left(\frac23\right)^2\).

The coefficient in front of the inverse tangent function depends on the constant factor in the denominator. Careful simplification is necessary to obtain the correct result.

🗺️ Solution Roadmap

Step-by-step Plan

›

Write the given integral.
Factor out \(9\) from the quadratic expression to prepare for completing the square.
Complete the square in \(x^2+\frac23x\).
Rewrite the denominator as a sum of two squares.
Apply the standard inverse tangent integral formula.
Simplify the argument and coefficient in the final expression.
Verify the answer by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 12 steps

Given —
\[I=\int\frac{1}{9x^2+6x+5}\,dx\]
Factor out \(9\) from the denominator.
We write
\[9x^2+6x+5=9\left(x^2+\frac23x+\frac59\right)\]
Therefore,
\[I=\frac19\int\frac{dx}{x^2+\frac23x+\frac59}\]
Complete the square
Consider the expression
\[x^2+\frac23x+\frac59\]
Take half the coefficient of \(x\):
\[\frac12\cdot\frac23=\frac13\]
Squaring this value gives
\[\left(\frac13\right)^2=\frac19\]
Add and subtract \(\frac19\) to complete the square:
\[ \begin{aligned} x^2+\frac23x+\frac59 &=x^2+\frac23x+\frac19+\frac59-\frac19\\[6pt] &=\left(x+\frac13\right)^2+\frac49 \end{aligned} \]
Since \(\frac49=\left(\frac23\right)^2\), we obtain
\[x^2+\frac23x+\frac59=\left(x+\frac13\right)^2+\left(\frac23\right)^2\]
Thus, the original denominator becomes
\[ \begin{aligned} 9x^2+6x+5 &=9\left[\left(x+\frac13\right)^2+\frac49\right]\\[6pt] &=9\left(x+\frac13\right)^2+4 \end{aligned} \]
Rewrite the integral
Substituting the completed-square form into the original integral gives
\[ I=\frac19\int \frac{dx}{\left(x+\frac13\right)^2+\left(\frac23\right)^2} \]
Apply the standard integral formula
We use
\[\int\frac{du}{u^2+a^2}=\frac1a\tan^{-1}\left(\frac ua\right)+C\]
Here,
\[u=x+\frac13,\quad a=\frac23\]
Since \(du=dx\), we obtain
\[ \begin{aligned} I &=\frac19\cdot\frac{1}{\frac23} \tan^{-1}\left( \frac{x+\frac13}{\frac23} \right)+C\\[8pt] &=\frac19\cdot\frac32 \tan^{-1}\left( \frac{3x+1}{2} \right)+C\\[8pt] &=\frac16\tan^{-1}\left(\frac{3x+1}{2}\right)+C \end{aligned} \]

💡 Answer

Final Answer

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\[ \boxed{ \displaystyle \int\frac{dx}{9x^2+6x+5} =\frac16\tan^{-1}\left(\frac{3x+1}{2}\right)+C } \]

🎯 Exam Significance

Exam Significance

›

Completing the square: The question tests the ability to convert a quadratic polynomial into a standard form suitable for integration.
Standard integral recognition: Students must identify the inverse tangent form after completing the square.
Coefficient management: The factor \(\frac19\) introduced by factoring out \(9\) must be retained throughout the calculation.
Algebraic accuracy: Correctly simplifying \(\frac{x+\frac13}{\frac23}\) to \(\frac{3x+1}{2}\) is essential.
Stepwise presentation: Showing the square completion, standard formula, and simplification supports a complete and logically sound solution.

Significance for Competitive Entrance Examinations

JEE Main and Advanced: This problem strengthens the use of completing the square and standard inverse tangent integrals.
NEET and other entrance examinations: Recognising the quadratic denominator pattern helps students solve integration questions efficiently.
Algebraic manipulation: Factoring constants and simplifying the completed-square form are useful skills in more complex integrals.
Coefficient accuracy: Tracking constant factors helps avoid incorrect numerical coefficients in objective answers.
Verification: Differentiating the final expression provides a reliable check of the result.

🚫 Common Mistakes

Common Mistakes

›

Forgetting the factor \(\frac19\) when factoring \(9\) out of the denominator.
Writing \(\left(x+\frac13\right)^2\) incorrectly as \(x^2+\frac19\), omitting the middle term.
Confusing \(\frac49\) with \(\left(\frac49\right)^2\).
Applying the standard inverse tangent formula without accounting for the factor \(\frac19\).
Writing \(\frac32\) instead of \(\frac16\) as the final coefficient.
Omitting the constant of integration \(C\).

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 6 points

Complete the square to transform the quadratic denominator into a standard form.
The identity \(x^2+\frac23x+\frac59=\left(x+\frac13\right)^2+\frac49\) is the key algebraic step.
The standard formula is \(\displaystyle\int\frac{du}{u^2+a^2}=\frac1a\tan^{-1}\left(\frac ua\right)+C\).
Retain the factor \(\frac19\) introduced when factoring \(9\) out of the denominator.
The correct coefficient in the final answer is \(\frac16\), not \(\frac32\).
Differentiate the final expression to verify the antiderivative.

📘 Concept & Theory

Theory and Concept

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This problem is based on completing the square and applying the standard integral formula for the inverse sine function. The expression inside the square root is a quadratic polynomial. By rewriting it as the difference of two squares, we can convert the integral into a standard form.

The standard formula is

\[ \int\frac{du}{\sqrt{a^2-u^2}} =\sin^{-1}\left(\frac{u}{a}\right)+C, \qquad a>0 \]

The key is to express \(7-6x-x^2\) in the form \(a^2-(x+b)^2\). Once this is achieved, the inverse sine formula can be applied directly.

🗺️ Solution Roadmap

Step-by-step Plan

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Write the given integral.
Rearrange the quadratic expression inside the square root.
Complete the square to express it as a difference of squares.
Identify the substitution required for the standard inverse sine formula.
Apply the formula and write the final answer.
Verify the result by differentiation.

✏️ Solution

Complete Solution

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Step-by-step Solution · 10 steps

Given —
\[I=\int\frac{dx}{\sqrt{7-6x-x^2}}\]
Rearrange the quadratic expression.
The expression under the square root is
\[7-6x-x^2\]
Rearranging the terms gives
\[7-6x-x^2=7-(x^2+6x)\]
Complete the square
To complete the square in \(x^2+6x\), take half the coefficient of \(x\) and square it:
\[\frac{6}{2}=3,\qquad 3^2=9\]
Add and subtract \(9\) inside the expression:
\[ \begin{aligned} 7-6x-x^2 &=7-(x^2+6x)\\[6pt] &=7-(x^2+6x+9-9)\\[6pt] &=7-\left[(x+3)^2-9\right]\\[6pt] &=7-(x+3)^2+9\\[6pt] &=16-(x+3)^2 \end{aligned} \]
Since \(16=4^2\), we obtain
\[7-6x-x^2=4^2-(x+3)^2\]
Substitute the completed-square form
Replacing the quadratic expression in the original integral gives
\[I=\int\frac{dx}{\sqrt{16-(x+3)^2}}\]
This matches the standard form
\[\int\frac{du}{\sqrt{a^2-u^2}}\]
Let
\[u=x+3,\quad du=dx,\quad a=4\]
Apply the standard integral formula
Using
\[\int\frac{du}{\sqrt{a^2-u^2}}=\sin^{-1}\left(\frac{u}{a}\right)+C\]
we get
\[\begin{aligned}I&=\sin^{-1}\left(\frac{u}{4}\right)+C\\[6pt] &=\sin^{-1}\left(\frac{x+3}{4}\right)+C\end{aligned}\]

💡 Answer

Final Answer

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\[ \boxed{ \displaystyle \int\frac{dx}{\sqrt{7-6x-x^2}} =\sin^{-1}\left(\frac{x+3}{4}\right)+C } \]

🎯 Exam Significance

Exam Significance

›

Completing the square: This is the central algebraic technique required to simplify the quadratic expression.
Recognition of standard forms: Students must recognise the inverse sine integral after rewriting the denominator.
Substitution: Identifying \(u=x+3\) makes the integral straightforward.
Stepwise presentation: Showing each algebraic transformation makes the solution clear and easier to assess.
Domain awareness: Determining where the radicand is positive helps ensure the integral is real and defined.

Significance for Competitive Entrance Examinations

JEE Main and Advanced: Completing the square is a useful technique for recognising standard integration patterns.
NEET and other entrance examinations: Quick identification of inverse trigonometric integral forms helps reduce unnecessary calculations.
Algebraic accuracy: The transformation from \(7-6x-x^2\) to \(16-(x+3)^2\) is a common source of sign errors.
Conceptual understanding: Connecting algebraic simplification with a standard integration formula supports the solution of more complex integrals.
Verification skills: Differentiation is an effective way to confirm an antiderivative.

🚫 Common Mistakes

Common Mistakes

›

Forgetting to add and subtract \(9\) when completing the square.
Writing \(7-(x^2+6x+9)\) without compensating for the added \(9\).
Using \(x-3\) instead of \(x+3\) in the completed-square expression.
Applying the inverse sine formula without identifying the correct value \(a=4\).
Forgetting the constant of integration \(C\).
Ignoring the restriction \(-7

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 6 points

Complete the square before applying the standard integral formula.
The essential identity is \(7-6x-x^2=16-(x+3)^2\).
Use the formula \(\displaystyle\int\frac{du}{\sqrt{a^2-u^2}}=\sin^{-1}\left(\frac{u}{a}\right)+C\).
Here, \(u=x+3\) and \(a=4\).
The integrand is real and defined for \(-7
Differentiate the final expression to verify the result.

📘 Concept & Theory

Theory and Concept

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This problem involves integrating a rational expression containing the square root of a quadratic polynomial. The key technique is to expand the product, complete the square, and apply a standard integral formula.

The relevant standard formula is

\[ \int\frac{du}{\sqrt{u^2-a^2}} =\ln\left|u+\sqrt{u^2-a^2}\right|+C \]

This formula applies where \(u^2-a^2>0\). We will transform the expression \((x-1)(x-2)\) into the form \(u^2-a^2\) by completing the square.

🗺️ Solution Roadmap

Step-by-step Plan

›

Expand the product \((x-1)(x-2)\).
Complete the square in the resulting quadratic expression.
Introduce a substitution to match the standard integral formula.
Apply the logarithmic integral formula.
Simplify the result and verify it by differentiation.
Identify the real domain of the original integrand.

✏️ Solution

Complete Solution

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Step-by-step Solution · 16 steps

Given —
\[I=\int\frac{dx}{\sqrt{(x-1)(x-2)}}\]
Expand the product
Using the algebraic identity \((a-b)(a-c)=a^2-(b+c)a+bc\), we get
\[\begin{aligned} (x-1)(x-2) &=x^2-2x-x+2\\ &=x^2-3x+2 \end{aligned}\]
Therefore,
\[I=\int\frac{dx}{\sqrt{x^2-3x+2}}\]
Complete the square.
Consider the quadratic expression
\[x^2-3x+2\]
Half the coefficient of \(x\) is
\[-\frac32\]
Squaring this gives
\[\left(-\frac32\right)^2=\frac94\]
Add and subtract \(\frac94\):
\[ \begin{aligned} x^2-3x+2 &=x^2-3x+\frac94+2-\frac94\\[6pt] &=\left(x-\frac32\right)^2-\frac14 \end{aligned} \]
Thus, the integral becomes
\[I=\int\frac{dx}{\sqrt{\left(x-\frac32\right)^2-\left(\frac12\right)^2}}\]
Use substitution
Let
\[u=x-\frac32\]
Differentiating with respect to \(x\),
\[du=dx\]
Substituting \(u=x-\frac32\) and \(dx=du\), we obtain
\[I=\int\frac{du}{\sqrt{u^2-\left(\frac12\right)^2}}\]
Apply the standard integral formula.
Recall that
\[\int\frac{du}{\sqrt{u^2-a^2}}=\ln\left|u+\sqrt{u^2-a^2}\right|+C\]
Here, \(a=\frac12\). Therefore,
\[I=\ln\left|u+\sqrt{u^2-\frac14}\right|+C\]
Substituting \(u=x-\frac32\), we get
\[I=\ln\left|x-\frac32+\sqrt{\left(x-\frac32\right)^2-\frac14}\right|+C\]
Since
\[\left(x-\frac32\right)^2-\frac14=(x-1)(x-2)\]
the result can be written as
\[I=\ln\left|x-\frac32+\sqrt{(x-1)(x-2)}\right|+C\]
Multiplying the expression inside the logarithm by \(2\) changes the logarithm only by the constant \(\ln 2\), which is absorbed into \(C\). Hence, an equivalent and simpler form is
\[I=\ln\left|2x-3+2\sqrt{(x-1)(x-2)}\right|+C\]

💡 Answer

Final Answer

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\[ \boxed{ \displaystyle \int\frac{dx}{\sqrt{(x-1)(x-2)}} =\ln\left|2x-3+2\sqrt{(x-1)(x-2)}\right|+C } \]

🎯 Exam Significance

Exam Significance

›

Completing the square: The question tests whether students can transform a quadratic expression into a standard form.
Standard integral formula: Students must recognise the logarithmic form involving \(\sqrt{u^2-a^2}\).
Substitution: The shift \(u=x-\frac32\) simplifies the integral without changing the differential.
Algebraic simplification: Correctly handling constants inside logarithms helps present equivalent answers.
Verification: Differentiation confirms the correctness of the antiderivative.

Significance for Competitive Entrance Examinations

JEE Main and Advanced: Completing the square and identifying standard integral forms are important integration techniques.
NEET and other entrance examinations: Recognising the logarithmic pattern can help solve objective questions efficiently.
Conceptual understanding: This example connects quadratic algebra, substitution, logarithms, and differentiation.
Domain analysis: Determining where the radicand is positive is important for real-valued integral expressions.

🚫 Common Mistakes

Common Mistakes

›

Expanding \((x-1)(x-2)\) incorrectly.
Forgetting to subtract \(\frac94\) after adding it while completing the square.
Using the inverse sine formula instead of the logarithmic formula for \(\sqrt{u^2-a^2}\).
Omitting the absolute value around the logarithm's argument when using the standard formula.
Assuming the integrand is real for every real \(x\).
Forgetting the constant of integration \(C\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

Expand the product before completing the square.
The key identity is \(x^2-3x+2=\left(x-\frac32\right)^2-\frac14\).
Use \(\displaystyle\int\frac{du}{\sqrt{u^2-a^2}}=\ln\left|u+\sqrt{u^2-a^2}\right|+C\).
The final answer can be expressed in more than one equivalent logarithmic form.
The real domain is \(x<1\) or \(x>2\).
Differentiate the answer to verify the result.

📘 Concept & Theory

Theory and Concept

›

This problem uses the method of completing the square to transform a quadratic expression inside a square root into the standard form required for inverse sine integration.

The relevant standard integral formula is

\[ \int\frac{du}{\sqrt{a^2-u^2}} =\sin^{-1}\left(\frac{u}{a}\right)+C, \qquad a>0 \]

To apply this formula, we must express \(8+3x-x^2\) as \(a^2-(x-b)^2\). Since the coefficient of \(x^2\) is negative, completing the square will produce a positive constant minus a perfect square.

🗺️ Solution Roadmap

Step-by-step Plan

›

Rearrange the quadratic expression under the square root.
Complete the square, paying attention to the sign of the linear term.
Express the result as the difference of two squares.
Identify the appropriate substitution for the standard inverse sine formula.
Simplify the inverse sine argument.
Verify the answer by differentiation and determine the real domain.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 11 steps

Given —
\[I=\int\frac{dx}{\sqrt{8+3x-x^2}}\]
Rearrange the quadratic expression.
Write the expression inside the square root as
\[8+3x-x^2=8-(x^2-3x)\]
We now complete the square in \(x^2-3x\).
Complete the square
Half the coefficient of \(x\) in \(x^2-3x\) is
\[-\frac32\]
Squaring it gives
\[\left(-\frac32\right)^2=\frac94\]
Add and subtract \(\frac94\):
\[ \begin{aligned} 8+3x-x^2 &=8-(x^2-3x)\\[6pt] &=8-\left(x^2-3x+\frac94-\frac94\right)\\[6pt] &=8+\frac94-\left(x^2-3x+\frac94\right)\\[6pt] &=\frac{32}{4}+\frac94-\left(x-\frac32\right)^2\\[6pt] &=\frac{41}{4}-\left(x-\frac32\right)^2 \end{aligned} \]
Therefore,
\[\boxed{8+3x-x^2=\left(\frac{\sqrt{41}}{2}\right)^2-\left(x-\frac32\right)^2}\]
Substitute the completed-square expression.
Substituting into the integral gives
\[I=\int\frac{dx}{\sqrt{\frac{41}{4}-\left(x-\frac32\right)^2}}\]
Comparing this with the standard form
\[\int\frac{du}{\sqrt{a^2-u^2}}\]
we identify
\[u=x-\frac32,\qquad du=dx,\qquad a=\frac{\sqrt{41}}{2}\]
Apply the standard inverse sine formula
Using
\[\int\frac{du}{\sqrt{a^2-u^2}}=\sin^{-1}\left(\frac{u}{a}\right)+C\]
we obtain
\[ \begin{aligned} I &=\sin^{-1}\left( \frac{x-\frac32}{\frac{\sqrt{41}}{2}} \right)+C\\[8pt] &=\sin^{-1}\left( \frac{2\left(x-\frac32\right)}{\sqrt{41}} \right)+C\\[8pt] &=\sin^{-1}\left(\frac{2x-3}{\sqrt{41}}\right)+C \end{aligned} \]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int\frac{dx}{\sqrt{8+3x-x^2}}=\sin^{-1}\left(\frac{2x-3}{\sqrt{41}}\right)+C}\]

🎯 Exam Significance

Exam Significance

›

Completing the square: This question tests the ability to transform a quadratic expression into a standard integration form.
Sign accuracy: The negative coefficient of \(x^2\) makes it essential to handle the linear term carefully.
Standard integral recognition: Students must recognise the inverse sine formula after completing the square.
Substitution: Identifying \(u=x-\frac32\) makes the integral straightforward.
Stepwise presentation: Showing the algebraic steps clearly supports a complete solution.

Significance for Competitive Entrance Examinations

JEE Main and Advanced: Completing the square is a useful technique for converting quadratic expressions into standard integral forms.
NEET and other entrance examinations: Recognising inverse trigonometric integral patterns helps solve questions efficiently.
Algebraic precision: Correctly identifying the centre shift \(x-\frac32\) avoids sign errors in the final answer.
Verification skills: Differentiation provides an effective way to check an antiderivative.

🚫 Common Mistakes

Common Mistakes

›

Changing \(8+3x-x^2\) incorrectly to \(8-3x-x^2\).
Writing \(\left(x+\frac32\right)^2\) instead of \(\left(x-\frac32\right)^2\).
Forgetting to add \(\frac94\) to the constant term when completing the square.
Using \(a=\frac{41}{4}\) instead of \(a=\frac{\sqrt{41}}{2}\).
Incorrectly simplifying the inverse sine argument.
Forgetting the constant of integration \(C\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

The key identity is \(8+3x-x^2=\frac{41}{4}-\left(x-\frac32\right)^2\).
Complete the square before applying the inverse sine formula.
Use \(\displaystyle\int\frac{du}{\sqrt{a^2-u^2}}=\sin^{-1}\left(\frac{u}{a}\right)+C\).
The correct inverse sine argument is \(\frac{2x-3}{\sqrt{41}}\).
The real domain is \(\displaystyle\frac{3-\sqrt{41}}{2}
Differentiate the result to confirm the answer.

📘 Concept & Theory

Theory and Concept

›

This problem involves integrating the reciprocal of the square root of a quadratic expression containing two parameters, \(a\) and \(b\). The main technique is to expand the product and complete the square.

We use the standard integral formula

\[\int\frac{du}{\sqrt{u^2-c^2}}=\ln\left|u+\sqrt{u^2-c^2}\right|+C\]

Assume that \(a\) and \(b\) are real constants. The derivation below applies when \(a\ne b\), on intervals where the original integrand is real and defined. The same final formula also works when \(a=b\), with the appropriate domain interpretation.

🗺️ Solution Roadmap

Step-by-step Plan

›

Expand \((x-a)(x-b)\).
Complete the square in the resulting quadratic expression.
Identify the substitution that reduces the integral to a standard logarithmic form.
Apply the standard integral formula.
Simplify the result and verify it by differentiation.
State the domain conditions for a real-valued integrand.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 18 steps

Given —
\[I=\int\frac{dx}{\sqrt{(x-a)(x-b)}}\]
Expand the product
Using the distributive property,
\[\begin{aligned}(x-a)(x-b)&=x^2-bx-ax+ab\\[6pt]&=x^2-(a+b)x+ab\end{aligned}\]
Therefore,
\[I=\int\frac{dx}{\sqrt{x^2-(a+b)x+ab}}\]
Complete the square
Consider the quadratic expression
\[x^2-(a+b)x+ab\]
Half the coefficient of \(x\) is
\[-\frac{a+b}{2}\]
Its square is
\[\frac{(a+b)^2}{4}\]
Add and subtract this quantity:
\[ \begin{aligned} x^2-(a+b)x+ab &=x^2-(a+b)x+\frac{(a+b)^2}{4}\\ &\quad+ab-\frac{(a+b)^2}{4}\\[6pt] &=\left(x-\frac{a+b}{2}\right)^2 +ab-\frac{(a+b)^2}{4} \end{aligned} \]
Simplify the constant term:
\[ \begin{aligned} ab-\frac{(a+b)^2}{4} &=ab-\frac{a^2+2ab+b^2}{4}\\[6pt] &=\frac{4ab-a^2-2ab-b^2}{4}\\[6pt] &=-\frac{(a-b)^2}{4} \end{aligned} \]
Hence,
\[\boxed{(x-a)(x-b)=\left(x-\frac{a+b}{2}\right)^2-\left(\frac{a-b}{2}\right)^2}\]
Substitute into the integral
Using the completed-square form, we get
\[I=\int\frac{dx}{\sqrt{\left(x-\frac{a+b}{2}\right)^2-\left(\frac{a-b}{2}\right)^2}}\]
Let
\[u=x-\frac{a+b}{2}\]
Since \(a\) and \(b\) are constants,
\[du=dx\]
The integral becomes
\[I=\int\frac{du}{\sqrt{u^2-\left(\frac{a-b}{2}\right)^2}}\]
Apply the standard integral formula
Recall that
\[\int\frac{du}{\sqrt{u^2-c^2}}=\ln\left|u+\sqrt{u^2-c^2}\right|+C\]
Taking \(c=\frac{|a-b|}{2}\), we obtain
\[I=\ln\left|u+\sqrt{u^2-\frac{(a-b)^2}{4}}\right|+C\]
Substituting \(u=x-\frac{a+b}{2}\),
\[I=\ln\left|x-\frac{a+b}{2}+\sqrt{\left(x-\frac{a+b}{2}\right)^2-\frac{(a-b)^2}{4}}\right|+C\]
From the completed-square identity,
\[\left(x-\frac{a+b}{2}\right)^2-\frac{(a-b)^2}{4}=(x-a)(x-b)\]
Therefore,
\[I=\ln\left|x-\frac{a+b}{2}+\sqrt{(x-a)(x-b)}\right|+C\]
Multiplying the expression inside the logarithm by \(2\) changes the result only by the constant \(\ln 2\), which is absorbed into \(C\). Thus, an equivalent form is
\[\boxed{I=\ln\left|2x-a-b+2\sqrt{(x-a)(x-b)}\right|+C}\]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int\frac{dx}{\sqrt{(x-a)(x-b)}}=\ln\left|2x-a-b+2\sqrt{(x-a)(x-b)}\right|+C}\]

🎯 Exam Significance

Exam Significance

›

Completing the square: The problem tests the ability to convert a quadratic polynomial into a standard form.
Parameter handling: Students must correctly manipulate the constants \(a\) and \(b\).
Standard integral recognition: The transformed integral matches the logarithmic formula involving \(\sqrt{u^2-c^2}\).
Algebraic simplification: Correctly simplifying the constant term to \(-\frac{(a-b)^2}{4}\) is essential.
Verification: Differentiation confirms the antiderivative.

Significance for Competitive Entrance Examinations

JEE Main and Advanced: This example develops fluency in parameter-based integration and completing the square.
NEET and other entrance examinations: Recognising standard integral forms helps simplify objective questions.
Algebraic reasoning: The identity links the roots \(a,b\) with the centre \(\frac{a+b}{2}\) and the squared half-difference \(\frac{(a-b)^2}{4}\).
Domain analysis: Understanding where the radicand is positive prevents invalid real-valued manipulations.

🚫 Common Mistakes

Common Mistakes

›

Expanding \((x-a)(x-b)\) with an incorrect sign.
Using \(\frac{a-b}{2}\) instead of its square when completing the square.
Forgetting the absolute value in the logarithmic formula.
Dropping constants inside the logarithm without explaining that the difference is absorbed into \(C\).
Assuming the integrand is real for every real value of \(x\).
Forgetting that \(a\) and \(b\) are constants, so \(du=dx\) for the chosen substitution.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

Expand and complete the square before integrating.
The key identity is
\[ (x-a)(x-b)=\left(x-\frac{a+b}{2}\right)^2-\left(\frac{a-b}{2}\right)^2 \]
Use the standard logarithmic integral formula for \(\sqrt{u^2-c^2}\).
The result can be written compactly as
\[ \ln\left|2x-a-b+2\sqrt{(x-a)(x-b)}\right|+C \]
For \(a\ne b\), the real domain is outside the interval between \(a\) and \(b\).
Differentiate the result to verify the answer.

📘 Concept & Theory

Theory and Concept

›

This problem is based on integration by substitution. The numerator, \(4x+1\), is the derivative of the quadratic expression \(2x^2+x-3\) inside the square root.

The standard integral formula is

\[ \int\frac{du}{\sqrt{u}}=2\sqrt{u}+C \]

More generally, if the numerator is the derivative of the expression under a square root, the substitution \(u=f(x)\) can reduce the integral to a simple power integral. This approach avoids unnecessary algebraic manipulation.

🗺️ Solution Roadmap

Step-by-step Plan

›

Identify the expression under the square root.
Choose it as the substitution variable.
Differentiate the substitution to obtain \(du\).
Rewrite the integral in terms of the new variable.
Integrate using the power rule.
Substitute back and verify the answer by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 9 steps

Given —
\[I=\int\frac{4x+1}{\sqrt{2x^2+x-3}}\,dx\]
Choose a suitable substitution
Since the derivative of \(2x^2+x-3\) is \(4x+1\), let
\[t=2x^2+x-3\]
Differentiating both sides with respect to \(x\),
\[\frac{dt}{dx}=4x+1\]
Therefore,
\[dt=(4x+1)\,dx\]
Rewrite the integral
Substituting \(t=2x^2+x-3\) and \((4x+1)\,dx=dt\), we obtain
\[I=\int\frac{dt}{\sqrt{t}}\]
Expressing the square root as a fractional power,
\[I=\int t^{-\frac12}\,dt\]
Apply the power rule of integration.
The power rule states
\[ \int t^n\,dt=\frac{t^{n+1}}{n+1}+C, \qquad n\ne-1 \]
Here, \(n=-\frac12\). Hence,
\[ \begin{aligned} I &=\frac{t^{-\frac12+1}}{-\frac12+1}+C\\[6pt] &=\frac{t^{\frac12}}{\frac12}+C\\[6pt] &=2t^{\frac12}+C\\[6pt] &=2\sqrt{t}+C \end{aligned} \]
Substitute back the original expression
Since
\[t=2x^2+x-3\]
we get
\[I=2\sqrt{2x^2+x-3}+C\]

💡 Answer

Final Answer

›

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

🎯 Exam Significance

Exam Significance

›

Substitution method: The question tests the ability to identify an inner function and its derivative.
Power rule: Students must correctly integrate \(t^{-1/2}\).
Chain rule connection: Understanding the relationship between differentiation and integration helps identify suitable substitutions.
Clear working: Showing \(dt=(4x+1)\,dx\) makes the substitution explicit and easy to follow.
Verification: Differentiation provides a quick check of the final answer.

Significance for Competitive Entrance Examinations

JEE Main and Advanced: Recognising the derivative of the expression under the radical is a useful strategy for direct integration questions.
NEET and other entrance examinations: A suitable substitution reduces the integral to a basic power rule.
Time efficiency: Identifying the numerator as the derivative of the inner expression avoids lengthy calculations.
Conceptual fluency: The problem reinforces the connection between the chain rule and reverse differentiation.

🚫 Common Mistakes

Common Mistakes

›

Forgetting that \(\frac{d}{dx}(2x^2+x-3)=4x+1\).
Writing \(dt=(2x+1)\,dx\) instead of \(dt=(4x+1)\,dx\).
Applying the power rule incorrectly to \(t^{-1/2}\).
Forgetting that dividing by \(\frac12\) is equivalent to multiplying by \(2\).
Failing to substitute \(t\) back into the final expression.
Omitting the constant of integration \(C\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

Choose the expression under the square root as the substitution variable when its derivative appears in the numerator.
The substitution is \(t=2x^2+x-3\), giving \(dt=(4x+1)\,dx\).
The reduced integral is \(\displaystyle\int t^{-1/2}\,dt=2\sqrt{t}+C\).
The correct antiderivative is \(2\sqrt{2x^2+x-3}+C\).
The real domain is \(x<-\frac32\) or \(x>1\).
Always verify the result by differentiation when possible.

📘 Concept & Theory

Theory and Concept

›

This problem can be solved by splitting the integrand into two simpler integrals. One integral involves the derivative of the expression under the square root, while the other is a standard logarithmic integral.

The relevant standard formulas are

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

and

\[ \int\frac{dx}{\sqrt{x^2-a^2}} =\ln\left|x+\sqrt{x^2-a^2}\right|+C \]

The first formula follows by substitution, whereas the second is a standard integral. Both are needed to evaluate the given expression.

🗺️ Solution Roadmap

Step-by-step Plan

›

Split the numerator \(x+2\) into two terms.
Separate the integral into two simpler integrals.
Evaluate the first integral using the substitution \(t=x^2-1\).
Evaluate the second integral using the standard logarithmic formula.
Add both results and verify the final expression by differentiation.
State the domain on which the original integrand is real and defined.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 14 steps

Given —
\[I=\int\frac{x+2}{\sqrt{x^2-1}}\,dx\]
Split the integral
Using the linearity property of integration,
\[\begin{aligned}I&=\int\frac{x}{\sqrt{x^2-1}}\,dx+2\int\frac{dx}{\sqrt{x^2-1}}\end{aligned}\]
Let
\[I_1=\int\frac{x}{\sqrt{x^2-1}}\,dx\]
and
\[I_2=\int\frac{dx}{\sqrt{x^2-1}}\]
Thus
\[I=I_1+2I_2\]
Evaluation of the First Integral
Consider
\[I_1=\int\frac{x}{\sqrt{x^2-1}}\,dx\]
Choose a substitution
Let
\[t=x^2-1\]
Differentiating both sides,
\[\frac{dt}{dx}=2x\]
Therefore,
\[dt=2x\,dx\]
or
\[x\,dx=\frac12\,dt\]
Substitute into the integral
Since \(t=x^2-1\), we have \(\sqrt{x^2-1}=\sqrt{t}\). Hence,
\[ \begin{aligned} I_1 &=\int\frac{x\,dx}{\sqrt{x^2-1}}\\[6pt] &=\frac12\int\frac{dt}{\sqrt{t}}\\[6pt] &=\frac12\int t^{-\frac12}\,dt \end{aligned} \]
Apply the power rule
Using
\[\int t^n\,dt=\frac{t^{n+1}}{n+1}+C,\quad n\ne-1\]
we obtain
\[ \begin{aligned} I_1 &=\frac12\cdot \frac{t^{\frac12}}{\frac12}+C\\[6pt] &=\sqrt{t}+C\\[6pt] &=\sqrt{x^2-1}+C \end{aligned} \]
Therefore,
\[\boxed{I_1=\sqrt{x^2-1}}\]
Evaluation of the Second Integral
Consider
\[I_2=\int\frac{dx}{\sqrt{x^2-1}}\]
Using the standard formula
\[\int\frac{dx}{\sqrt{x^2-a^2}}=\ln\left|x+\sqrt{x^2-a^2}\right|+C\]
and taking \(a=1\), we obtain
\[\boxed{I_2=\ln\left|x+\sqrt{x^2-1}\right|+C}\]
Add Both Integrals
Recall that
\[I=I_1+2I_2\]
Substituting the results,
\[ \begin{aligned} I &=\sqrt{x^2-1} +2\ln\left|x+\sqrt{x^2-1}\right|+C \end{aligned} \]

💡 Answer

Final Answer

›

\[ \boxed{ \displaystyle \int\frac{x+2}{\sqrt{x^2-1}}\,dx = \sqrt{x^2-1} +2\ln\left|x+\sqrt{x^2-1}\right|+C } \]

🎯 Exam Significance

Exam Significance

›

Splitting integrals: The problem tests the linearity property of integration.
Substitution: The first integral requires recognising the derivative of \(x^2-1\).
Standard formulas: The second integral tests knowledge of the logarithmic form involving \(\sqrt{x^2-a^2}\).
Algebraic accuracy: The correct expression \(\sqrt{x^2-1}\) must be retained in the final answer.
Verification: Differentiation confirms that both parts combine to reproduce the original integrand.

Significance for Competitive Entrance Examinations

JEE Main and Advanced: This problem combines substitution with a standard logarithmic integral.
NEET and other entrance examinations: Recognising the two integral patterns helps simplify calculations.
Time efficiency: Splitting the numerator avoids unnecessary algebraic manipulation.
Error detection: Differentiation is useful for identifying sign mistakes in proposed antiderivatives.

🚫 Common Mistakes

Common Mistakes

›

Writing \(\sqrt{x^2+1}\) instead of \(\sqrt{x^2-1}\).
Forgetting the coefficient \(2\) multiplying the second integral.
Applying the power rule without accounting for the factor \(\frac12\) from the substitution.
Omitting the absolute value in the logarithmic formula.
Forgetting the constant of integration \(C\).
Ignoring the domain restriction \(x<-1\) or \(x>1\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

Split the numerator \(x+2\) into \(x\) and \(2\).
Use \(t=x^2-1\) to evaluate the first integral.
Use the standard logarithmic formula for the second integral.
The correct first term is \(\sqrt{x^2-1}\), not \(\sqrt{x^2+1}\).
The real domain is \(x<-1\) or \(x>1\).
Verify the final answer by differentiation.

📘 Concept & Theory

Theory and Concept

›

This problem combines algebraic decomposition, substitution, completing the square, and inverse tangent integration. The numerator is a linear expression, while the denominator is a quadratic polynomial.

The key idea is to express the numerator as a linear combination of the derivative of the denominator and a constant:

\[ 5x-2=A(6x+2)+B \]

This allows us to split the original integral into two parts:

\[ \int\frac{f'(x)}{f(x)}\,dx \qquad\text{and}\qquad \int\frac{dx}{f(x)} \]

The first part produces a logarithm, while the second can be evaluated by completing the square and using the standard formula

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

🗺️ Solution Roadmap

Step-by-step Plan

›

Differentiate the quadratic denominator.
Express the numerator as a multiple of this derivative plus a constant.
Split the integral into two simpler integrals.
Evaluate the first integral using logarithmic integration.
Complete the square in the denominator of the second integral.
Apply the inverse tangent formula.
Combine the results and verify by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 30 steps

Given —
\[I=\int\frac{5x-2}{1+2x+3x^2}\,dx\]
Let
\[f(x)=1+2x+3x^2\]
Differentiating,
\[f'(x)=6x+2\]
Express the numerator in terms of the derivative
Assume
\[5x-2=A(6x+2)+B\]
Expanding the right-hand side,
\[5x-2=6Ax+2A+B\]
Comparing the coefficients of \(x\),
\[6A=5\]
Therefore,
\[A=\frac56\]
Comparing the constant terms,
\[2A+B=-2\]
Substituting \(A=\frac56\),
\[ \begin{aligned} \frac53+B&=-2\\[4pt] B&=-2-\frac53\\[4pt] B&=-\frac{11}{3} \end{aligned} \]
Thus,
\[\boxed{5x-2=\frac56(6x+2)-\frac{11}{3}}\]
Split the integral
Substituting the decomposition into the original integral,
\[ \begin{aligned} I &=\int\frac{\frac56(6x+2)-\frac{11}{3}} {1+2x+3x^2}\,dx\\[8pt] &=\frac56\int\frac{6x+2}{1+2x+3x^2}\,dx -\frac{11}{3}\int\frac{dx}{1+2x+3x^2} \end{aligned} \]
Let
\[I_1=\frac56\int\frac{6x+2}{1+2x+3x^2}\,dx\]
and
\[I_2=-\frac{11}{3}\int\frac{dx}{1+2x+3x^2}\]
Therefore,
\[I=I_1+I_2\]
Evaluation of the First Integral
Consider
\[I_1=\frac56\int\frac{6x+2}{1+2x+3x^2}\,dx\]
Use substitution
Let
\[t=1+2x+3x^2\]
Differentiating,
\[dt=(6x+2)\,dx\]
Rewrite and integrate
Substituting \(t\) and \(dt\),
\[ \begin{aligned} I_1 &=\frac56\int\frac{dt}{t}\\[6pt] &=\frac56\ln|t|+C\\[6pt] &=\frac56\ln|1+2x+3x^2|+C \end{aligned} \]
The quadratic \(1+2x+3x^2\) is always positive for real \(x\), since its discriminant is
\[\Delta=2^2-4(3)(1)=4-12=-8<0\]
and its leading coefficient is positive. Therefore, the absolute value can be omitted:
\[\boxed{I_1=\frac56\ln(1+2x+3x^2)}\]
Evaluation of the Second Integral
Consider
\[I_2=-\frac{11}{3}\int\frac{dx}{1+2x+3x^2}\]
Factor out \(3\) from the quadratic expression.
\[1+2x+3x^2=3\left(x^2+\frac23x+\frac13\right)\]
Hence,
\[I_2=-\frac{11}{9}\int\frac{dx}{x^2+\frac23x+\frac13}\]
Complete the square
Half the coefficient of \(x\) is
\[\frac12\cdot\frac23=\frac13\]
Its square is
\[\left(\frac13\right)^2=\frac19\]
Add and subtract \(\frac19\):
\[ \begin{aligned} x^2+\frac23x+\frac13 &=x^2+\frac23x+\frac19+\frac13-\frac19\\[6pt] &=\left(x+\frac13\right)^2+\frac29\\[6pt] &=\left(x+\frac13\right)^2+ \left(\frac{\sqrt2}{3}\right)^2 \end{aligned} \]
Therefore,
\[ I_2=-\frac{11}{9} \int\frac{dx}{ \left(x+\frac13\right)^2+ \left(\frac{\sqrt2}{3}\right)^2} \]
Apply the inverse tangent formula
\[\int\frac{dx}{(x+h)^2+a^2}=\frac1a\tan^{-1}\left(\frac{x+h}{a}\right)+C\]
Here,
\[h=\frac13,\qquad a=\frac{\sqrt2}{3}\]
Thus,
\[ \begin{aligned} I_2 &=-\frac{11}{9}\cdot\frac{3}{\sqrt2} \tan^{-1}\left( \frac{x+\frac13}{\frac{\sqrt2}{3}} \right)+C\\[8pt] &=-\frac{11}{3\sqrt2} \tan^{-1}\left(\frac{3x+1}{\sqrt2}\right)+C \end{aligned} \]
Hence,
\[ \boxed{ I_2=-\frac{11}{3\sqrt2} \tan^{-1}\left(\frac{3x+1}{\sqrt2}\right) } \]
Combine Both Results
Since,
\[I=I_1+I_2\]
substituting the two results gives
\[ \begin{aligned} I &=\frac56\ln(1+2x+3x^2)\\[4pt] &\quad-\frac{11}{3\sqrt2} \tan^{-1}\left(\frac{3x+1}{\sqrt2}\right)+C \end{aligned} \]

💡 Answer

Final Answer

›

\[ \boxed{ \begin{aligned} \displaystyle \int\frac{5x-2}{1+2x+3x^2}\,dx &=\frac56\ln(1+2x+3x^2)\\[4pt] &\quad-\frac{11}{3\sqrt2} \tan^{-1}\left(\frac{3x+1}{\sqrt2}\right)+C \end{aligned} } \]

🎯 Exam Significance

Exam Significance

›

Numerator decomposition: The method tests the ability to express a linear numerator in terms of the derivative of the denominator and a constant.
Logarithmic integration: The first part follows the standard formula \(\int\frac{f'(x)}{f(x)}\,dx=\ln|f(x)|+C\).
Completing the square: The second part requires rewriting a quadratic expression in a standard form.
Coefficient accuracy: Students must retain all factors introduced when factoring constants from the denominator.
Verification: Differentiation confirms the correctness of the combined result.

Significance for Competitive Entrance Examinations

JEE Main and Advanced: This problem combines algebraic decomposition with logarithmic and inverse tangent integrals.
NEET and other entrance examinations: Recognising the derivative of the denominator helps simplify integration questions.
Efficient calculation: Separating the numerator into suitable terms reduces a complicated integral to two standard forms.
Error checking: Differentiation can identify missing factors in the inverse tangent coefficient.

🚫 Common Mistakes

Common Mistakes

›

Using an incorrect value of \(A\) or \(B\) in the numerator decomposition.
Omitting the coefficient \(-\frac{11}{3}\) when splitting the integral.
Forgetting the factor \(\frac13\) introduced when factoring \(3\) from the denominator.
Using \(-\frac{11}{\sqrt2}\) instead of the correct coefficient \(-\frac{11}{3\sqrt2}\).
Completing the square incorrectly or using the wrong inverse tangent argument.
Forgetting the constant of integration \(C\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

Express the numerator as a multiple of the derivative of the denominator plus a constant.
The first integral gives a logarithmic term.
The second integral gives an inverse tangent term after completing the square.
Retain the factor \(\frac13\) when factoring \(3\) from the quadratic denominator.
The correct inverse tangent coefficient is \(-\frac{11}{3\sqrt2}\).
Differentiate the complete answer to verify the result.

📘 Concept & Theory

Theory and Concept

›

This problem combines algebraic decomposition, substitution, completing the square, and logarithmic integration. The numerator is a linear expression, and the expression inside the square root is a quadratic polynomial.

The key idea is to express the numerator as a linear combination of the derivative of the quadratic expression and a constant. This separates the integral into two standard forms.

The first standard formula is

\[ \int\frac{f'(x)}{\sqrt{f(x)}}\,dx =2\sqrt{f(x)}+C \]

The second standard formula is

\[ \int\frac{du}{\sqrt{u^2-a^2}} =\ln\left|u+\sqrt{u^2-a^2}\right|+C \]

The second formula becomes applicable after completing the square in the quadratic expression.

🗺️ Solution Roadmap

Step-by-step Plan

›

Expand the product \((x-5)(x-4)\).
Express the numerator in terms of the derivative of the quadratic denominator.
Split the original integral into two parts.
Evaluate the first part using substitution.
Complete the square and evaluate the second part using the logarithmic formula.
Combine the results and verify by differentiation.
Determine the real domain of the integrand.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 34 steps

Given —
\[I=\int\frac{6x+7}{\sqrt{(x-5)(x-4)}}\,dx\]
Expand the product
\[ \begin{aligned} (x-5)(x-4) &=x^2-4x-5x+20\\[6pt] &=x^2-9x+20 \end{aligned} \]
Hence, the integral becomes
\[I=\int\frac{6x+7}{\sqrt{x^2-9x+20}}\,dx\]
Express the numerator in terms of the derivative
Let
\[f(x)=x^2-9x+20\]
Differentiating,
\[f'(x)=2x-9\]
Assume that
\[6x+7=A(2x-9)+B\]
Expanding the right-hand side,
\[6x+7=2Ax-9A+B\]
Comparing the coefficients of \(x\),
\[2A=6\]
Therefore,
\[A=3\]
Comparing the constant terms,
\[-9A+B=7\]
Substituting \(A=3\),
\[ \begin{aligned} -27+B&=7\\[4pt] B&=34 \end{aligned} \]
Thus,
\[\boxed{6x+7=3(2x-9)+34}\]
Split the integral
Substituting the decomposition,
\[ \begin{aligned} I &=\int\frac{3(2x-9)+34}{\sqrt{x^2-9x+20}}\,dx\\[8pt] &=3\int\frac{2x-9}{\sqrt{x^2-9x+20}}\,dx +34\int\frac{dx}{\sqrt{x^2-9x+20}} \end{aligned} \]
Let
\[I_1=3\int\frac{2x-9}{\sqrt{x^2-9x+20}}\,dx\]
and
\[I_2=34\int\frac{dx}{\sqrt{x^2-9x+20}}\]
Therefore,
\[I=I_1+I_2\]
Evaluation of the First Integral
Consider
\[I_1=3\int\frac{2x-9}{\sqrt{x^2-9x+20}}\,dx\]
Choose a substitution
Let
\[t=x^2-9x+20\]
Differentiating,
\[dt=(2x-9)\,dx\]
Rewrite the integral
Substituting \(t\) and \(dt\),
\[ \begin{aligned} I_1 &=3\int\frac{dt}{\sqrt{t}}\\[6pt] &=3\int t^{-\frac12}\,dt \end{aligned} \]
Apply the power rule
Using
\[\int t^n\,dt=\frac{t^{n+1}}{n+1}+C,\quad n\ne-1\]
we obtain
\[ \begin{aligned} I_1 &=3\cdot\frac{t^{\frac12}}{\frac12}+C\\[6pt] &=6\sqrt{t}+C\\[6pt] &=6\sqrt{x^2-9x+20}+C \end{aligned} \]
Therefore,
\[\boxed{I_1=6\sqrt{x^2-9x+20}}\]
Evaluation of the Second Integral
Consider
\[I_2=34\int\frac{dx}{\sqrt{x^2-9x+20}}\]
Complete the square
The quadratic expression is
\[x^2-9x+20\]
Half the coefficient of \(x\) is
\[-\frac92\]
Squaring this value gives
\[\left(-\frac92\right)^2=\frac{81}{4}\]
Add and subtract \(\frac{81}{4}\):
\[ \begin{aligned} x^2-9x+20 &=x^2-9x+\frac{81}{4}+20-\frac{81}{4}\\[6pt] &=\left(x-\frac92\right)^2+\frac{80-81}{4}\\[6pt] &=\left(x-\frac92\right)^2-\frac14\\[6pt] &=\left(x-\frac92\right)^2-\left(\frac12\right)^2 \end{aligned} \]
Hence,
\[I_2=34\int\frac{dx}{\sqrt{\left(x-\frac92\right)^2-\left(\frac12\right)^2}}\]
Apply the standard logarithmic formula
\[ \int\frac{du}{\sqrt{u^2-a^2}} =\ln\left|u+\sqrt{u^2-a^2}\right|+C \]
Let
\[u=x-\frac92,\quad du=dx,\quad a=\frac12\]
Applying the formula,
\[ I_2=34\ln\left| x-\frac92+\sqrt{\left(x-\frac92\right)^2-\frac14} \right|+C \]
Since
\[\left(x-\frac92\right)^2-\frac14=x^2-9x+20\]
we obtain
\[ I_2=34\ln\left| x-\frac92+\sqrt{x^2-9x+20} \right|+C \]
Multiplying the logarithm's argument by \(2\) changes the result only by the constant \(\ln 2\), which is absorbed into \(C\). Therefore, an equivalent form is
\[\boxed{I_2=34\ln\left|2x-9+2\sqrt{x^2-9x+20}\right|}\]
Combine Both Results
Since
\[I=I_1+I_2\]
substituting both results gives
\[ \begin{aligned} I &=6\sqrt{x^2-9x+20}\\[4pt] &\quad+34\ln\left| x-\frac92+\sqrt{x^2-9x+20} \right|+C \end{aligned} \]

💡 Answer

Final Answer

›

\[ \boxed{ \begin{aligned} \displaystyle \int\frac{6x+7}{\sqrt{(x-5)(x-4)}}\,dx &=6\sqrt{x^2-9x+20}\\[4pt] &\quad+34\ln\left| 2x-9+2\sqrt{x^2-9x+20} \right|+C \end{aligned} } \]

🎯 Exam Significance

Exam Significance

›

Algebraic decomposition: The question tests the ability to express a linear numerator in terms of the derivative of a quadratic expression.
Substitution: The first integral follows directly from the derivative of the expression under the square root.
Completing the square: The second integral requires transforming the quadratic into the form \(u^2-a^2\).
Standard formulas: Students must apply the logarithmic integral correctly.
Verification: Differentiating the combined answer confirms its correctness.

Significance for Competitive Entrance Examinations

JEE Main and Advanced: This problem combines algebraic decomposition with substitution and logarithmic integration.
NEET and other entrance examinations: Recognising the derivative of the quadratic expression helps simplify the integral.
Efficient problem-solving: Splitting the numerator reduces a complex-looking integral to two standard forms.
Algebraic accuracy: Correctly retaining the coefficient \(34\) and the square-root expression is essential.

🚫 Common Mistakes

Common Mistakes

›

Expanding \((x-5)(x-4)\) incorrectly.
Using \(B=27\) instead of \(B=34\) in the numerator decomposition.
Forgetting the factor \(3\) in the first integral.
Completing the square incorrectly and obtaining the wrong constant \(-\frac14\).
Omitting the absolute value in the logarithmic formula.
Forgetting that the original integrand is real only for \(x<4\) or \(x>5\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 6 points

The quadratic expression is \(x^2-9x+20\).
The numerator decomposition is \(6x+7=3(2x-9)+34\).
The first integral gives \(6\sqrt{x^2-9x+20}\).
The second integral produces a logarithmic term after completing the square.
The real domain is \(x<4\) or \(x>5\).
Differentiate the final expression to verify the result.

📘 Concept & Theory

Theory and Concept

›

This problem can be solved by expressing the numerator as a linear combination of the derivative of the expression inside the square root and a constant. We then split the original integral into two standard integrals.

The main concepts used are:

Integration by substitution: Used when the numerator contains the derivative of the expression under the square root.
Completing the square: Used to transform the quadratic expression into the standard form \(a^2-u^2\).
Standard integral formula:

\[ \int \frac{du}{\sqrt{a^2-u^2}} =\sin^{-1}\left(\frac{u}{a}\right)+C \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Express \(x+2\) in terms of the derivative of \(4x-x^2\) and a constant.
Split the integral into two simpler integrals.
Evaluate the first integral using substitution.
Complete the square in the second integral and apply the inverse-sine formula.
Combine the results and verify the answer by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 19 steps

Express the numerator appropriately
Let
\[x+2=A\frac{d}{dx}(4x-x^2)+B\]
Since
\[\frac{d}{dx}(4x-x^2)=4-2x,\]
we obtain
\[x+2=A(4-2x)+B\]

\[x+2=-2Ax+4A+B\]
Comparing the coefficients of \(x\) and the constant terms gives
\[-2A=1\implies A=-\frac12\]

\[4A+B=2\]
Substituting \(A=-\frac12\):
\[4\left(-\frac12\right)+B=2\]

\[-2+B=2\implies B=4\]
Therefore,
\[x+2=-\frac12(4-2x)+4\]
Split the integral
Substituting the decomposition into the original integral,
\[ \begin{aligned} I&=\int\frac{x+2}{\sqrt{4x-x^2}}\,dx\\[5pt] &= -\frac12\int\frac{4-2x}{\sqrt{4x-x^2}}\,dx +4\int\frac{dx}{\sqrt{4x-x^2}} \end{aligned} \]
Let
\[I_1=-\frac12\int\frac{4-2x}{\sqrt{4x-x^2}}\,dx\]
and
\[I_2=4\int\frac{dx}{\sqrt{4x-x^2}}.\]
Evaluate the first integral.
For \(I_1\), use the substitution
\[t=4x-x^2\]
Differentiating with respect to \(x\),
\[dt=(4-2x)\,dx\]
Therefore,
\[ \begin{aligned} I_1 &=-\frac12\int\frac{dt}{\sqrt{t}}\\[5pt] &=-\frac12\int t^{-1/2}\,dt\\[5pt] &=-\frac12\left(\frac{t^{1/2}}{1/2}\right)\\[5pt] &=-\sqrt{t} \end{aligned} \]
Substituting \(t=4x-x^2\), we get
\[\boxed{I_1=-\sqrt{4x-x^2}}\]
Evaluate the second integral
\[I_2=4\int\frac{dx}{\sqrt{4x-x^2}}\]
Complete the square in the denominator:
\[ \begin{aligned} 4x-x^2 &=-(x^2-4x)\\[5pt] &=-\left(x^2-4x+4-4\right)\\[5pt] &=4-(x-2)^2 \end{aligned} \]
Hence,
\[I_2=4\int\frac{dx}{\sqrt{4-(x-2)^2}}\]
Using the standard formula
\[\int\frac{du}{\sqrt{a^2-u^2}}=\sin^{-1}\left(\frac{u}{a}\right)+C,\]
take \(u=x-2\), so that \(du=dx\), and \(a=2\). Thus,
\[\begin{aligned}I_2&=4\sin^{-1}\left(\frac{x-2}{2}\right)\end{aligned}\]
Therefore,
\[\boxed{I_2=4\sin^{-1}\left(\frac{x-2}{2}\right)}\]
Combine both results
Since \(I=I_1+I_2\),
\[\begin{aligned}I&=-\sqrt{4x-x^2}+4\sin^{-1}\left(\frac{x-2}{2}\right)+C\end{aligned}\]

💡 Answer

Final Answer

›

\[ \boxed{ \begin{aligned} \int\frac{x+2}{\sqrt{4x-x^2}}\,dx ={}&-\sqrt{4x-x^2}\\ &+4\sin^{-1}\left(\frac{x-2}{2}\right)+C \end{aligned} } \]

🎯 Exam Significance

Exam Significance

›

This problem tests the method of decomposing a linear numerator into a multiple of the derivative of a quadratic expression plus a constant.
It combines substitution with the standard inverse-sine integral formula.
Showing the coefficient comparison and completing-square steps clearly can help secure method marks.
Always use the numerator exactly as given in the question before determining the constants.

Significance for Competitive Entrance Examinations

The same decomposition method applies to many integrals of the form \(\int \frac{mx+n}{\sqrt{ax^2+bx+c}}\,dx\).
Recognising the derivative of the quadratic expression often reduces the problem to a simple substitution.
Completing the square helps identify whether the remaining integral leads to an inverse-sine or logarithmic expression.
Differentiating the final expression is an efficient way to check signs and numerical coefficients.

🚫 Common Mistakes

Common Mistakes

›

Writing \(x-2\) instead of the given numerator \(x+2\).
Incorrectly comparing constant terms while determining \(A\) and \(B\).
Forgetting the factor \(4\) multiplying the second integral.
Using \(\sin^{-1}(x/2)\) instead of \(\sin^{-1}((x-2)/2)\) after completing the square.
Omitting the constant of integration \(C\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 4 points

Express the numerator in terms of the derivative of the quadratic expression under the radical.
Use substitution for the derivative-containing part.
Complete the square to obtain the standard form \(a^2-u^2\).
Verify the final answer by differentiation.

📘 Concept & Theory

Theory and Concept

›

This integral can be evaluated by expressing the numerator as a linear combination of the derivative of the expression inside the square root and a constant. This separates the integral into two standard forms.

The main concepts used are:

Decomposition of the numerator: Express the numerator in terms of the derivative of the quadratic expression.
Integration by substitution: Use the substitution \(t=x^2+2x+3\) for the first part.
Completing the square: Rewrite the quadratic expression as \((x+1)^2+2\).
Standard logarithmic integral:

\[ \int\frac{du}{\sqrt{u^2+a^2}} =\ln\left|u+\sqrt{u^2+a^2}\right|+C \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Express \(x+2\) as \(A(2x+2)+B\).
Determine the constants \(A\) and \(B\) by comparing coefficients.
Split the original integral into two parts.
Evaluate the first part using substitution.
Complete the square and evaluate the second part using the standard logarithmic formula.
Combine the results and verify the answer by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 22 steps

Express the numerator appropriately
Let
\[x+2=A\frac{d}{dx}(x^2+2x+3)+B\]
Since
\[\frac{d}{dx}(x^2+2x+3)=2x+2,\]
we obtain
\[x+2=A(2x+2)+B\]

\[x+2=2Ax+2A+B\]
Comparing the coefficients of \(x\),
\[2A=1\]

\[A=\frac12\]
Comparing the constant terms,
\[2A+B=2\]
Substituting \(A=\frac12\),
\[2\left(\frac12\right)+B=2\]

\[1+B=2\]

\[B=1\]
Therefore,
\[\boxed{x+2=\frac12(2x+2)+1}\]
Split the integral
Substituting the decomposition into the original integral,
\[ \begin{aligned} I &=\int\frac{x+2}{\sqrt{x^2+2x+3}}\,dx\\[6pt] &=\int\frac{\frac12(2x+2)+1} {\sqrt{x^2+2x+3}}\,dx\\[6pt] &=\frac12\int\frac{2x+2}{\sqrt{x^2+2x+3}}\,dx +\int\frac{dx}{\sqrt{x^2+2x+3}} \end{aligned} \]
Let
\[I_1=\frac12\int\frac{2x+2}{\sqrt{x^2+2x+3}}\,dx\]
and
\[I_2=\int\frac{dx}{\sqrt{x^2+2x+3}}.\]
Evaluate the first integral
\[I_1=\frac12\int\frac{2x+2}{\sqrt{x^2+2x+3}}\,dx\]
Use the substitution
\[t=x^2+2x+3\]
Differentiating with respect to \(x\),
\[dt=(2x+2)\,dx\]
Therefore,
\[ \begin{aligned} I_1 &=\frac12\int\frac{dt}{\sqrt{t}}\\[6pt] &=\frac12\int t^{-1/2}\,dt\\[6pt] &=\frac12\left(\frac{t^{1/2}}{1/2}\right)\\[6pt] &=\sqrt{t} \end{aligned} \]
Substituting \(t=x^2+2x+3\), we get
\[\boxed{I_1=\sqrt{x^2+2x+3}}\]
Evaluate the second integral
\[I_2=\int\frac{dx}{\sqrt{x^2+2x+3}}\]
Complete the square in the denominator:
\[ \begin{aligned} x^2+2x+3 &=x^2+2x+1+2\\[5pt] &=(x+1)^2+2\\[5pt] &=(x+1)^2+(\sqrt2)^2 \end{aligned} \]
Hence,
\[I_2=\int\frac{dx}{\sqrt{(x+1)^2+(\sqrt2)^2}}\]
Recall the standard integral formula:
\[\int\frac{du}{\sqrt{u^2+a^2}}=\ln\left|u+\sqrt{u^2+a^2}\right|+C\]
Let
\[u=x+1,\quad du=dx,\quad a=\sqrt2.\]
Applying the formula,
\[ \begin{aligned} I_2 &=\ln\left|u+\sqrt{u^2+2}\right|+C\\[5pt] &=\ln\left|x+1+\sqrt{(x+1)^2+2}\right|+C \end{aligned} \]
Since \((x+1)^2+2=x^2+2x+3\),
\[\boxed{I_2=\ln\left|x+1+\sqrt{x^2+2x+3}\right|}\]
Combine both results
Since \(I=I_1+I_2\),
\[ \begin{aligned} I={}&\sqrt{x^2+2x+3}\\ &+\ln\left|x+1+\sqrt{x^2+2x+3}\right|+C \end{aligned} \]

💡 Answer

Final Answer

›

\[ \boxed{ \begin{aligned} \int\frac{x+2}{\sqrt{x^2+2x+3}}\,dx ={}&\sqrt{x^2+2x+3}\\ &+\ln\left|x+1+\sqrt{x^2+2x+3}\right|+C \end{aligned} } \]

🎯 Exam Significance

Exam Significance

›

This problem demonstrates how to decompose a linear numerator into a multiple of the derivative of a quadratic expression plus a constant.
It combines substitution with the standard logarithmic integral formula.
Showing the coefficient comparison, substitution, and completing-square steps clearly helps establish the method and supports method marks.
Writing the constant of integration \(C\) is essential in an indefinite integral.

Significance for Competitive Entrance Examinations

The decomposition technique is useful for integrals involving a linear numerator and a quadratic expression under a square root.
Recognising the derivative of the quadratic expression simplifies one part of the integral immediately.
Completing the square helps identify the correct standard formula for the remaining part.
Differentiation provides a quick and reliable way to check the final result.

🚫 Common Mistakes

Common Mistakes

›

Incorrectly calculating the constants \(A\) and \(B\). The correct values are \(A=\frac12\) and \(B=1\).
Forgetting the factor \(\frac12\) in the first integral after decomposition.
Applying the logarithmic formula without first completing the square.
Writing \(\sqrt{x^2+2x+3}\) incorrectly after substituting back.
Omitting the constant of integration \(C\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

Decompose the numerator using the derivative of the expression under the square root.
Use substitution for the derivative-containing part of the integral.
Complete the square to transform the remaining integral into the standard form \(\int\frac{du}{\sqrt{u^2+a^2}}\).
Verify the answer by differentiating the complete antiderivative.
The final result is valid for every real \(x\), because the quadratic expression under the square root is always positive.

📘 Concept & Theory

Theory and Concept

›

This integral can be evaluated by expressing the numerator as a linear combination of the derivative of the denominator and a constant. The resulting integral is split into two parts: one involving a logarithm and another involving the difference of two squares.

The main concepts used are:

Numerator decomposition: Express the numerator in terms of the derivative of the denominator.
Substitution: Use the derivative of the denominator to evaluate the first integral.
Completing the square: Rewrite the quadratic denominator in the form \(u^2-a^2\).
Standard logarithmic formula:

\[ \int\frac{du}{u^2-a^2} =\frac{1}{2a}\ln\left|\frac{u-a}{u+a}\right|+C \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Express \(x+3\) in terms of \(2x-2\) and a constant.
Determine the constants by comparing coefficients.
Split the original integral into two simpler integrals.
Evaluate the first integral using substitution.
Complete the square in the second integral and apply the standard formula.
Combine the results and verify the answer by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 21 steps

Express the numerator appropriately
Let
\[x+3=A\frac{d}{dx}(x^2-2x-5)+B\]
Since
\[\frac{d}{dx}(x^2-2x-5)=2x-2,\]
we obtain
\[x+3=A(2x-2)+B\]

\[x+3=2Ax-2A+B\]
Comparing the coefficients of \(x\),
\[2A=1\implies A=\frac12\]
Comparing the constant terms,
\[-2A+B=3\]
Substituting \(A=\frac12\),
\[-2\left(\frac12\right)+B=3\]

\[-1+B=3\]

\[B=4\]
Therefore, the correct decomposition is
\[\boxed{x+3=\frac12(2x-2)+4}\]
Split the integral
Substituting the correct decomposition,
\[ \begin{aligned} I &=\int\frac{x+3}{x^2-2x-5}\,dx\\[6pt] &=\int\frac{\frac12(2x-2)+4}{x^2-2x-5}\,dx\\[6pt] &=\frac12\int\frac{2x-2}{x^2-2x-5}\,dx +4\int\frac{dx}{x^2-2x-5} \end{aligned} \]
Let
\[I_1=\frac12\int\frac{2x-2}{x^2-2x-5}\,dx\]
and
\[I_2=4\int\frac{dx}{x^2-2x-5}.\]
Evaluate the first integral
\[I_1=\frac12\int\frac{2x-2}{x^2-2x-5}\,dx\]
Use the substitution
\[t=x^2-2x-5\]
Differentiating,
\[dt=(2x-2)\,dx\]
Therefore,
\[ \begin{aligned} I_1 &=\frac12\int\frac{dt}{t}\\[6pt] &=\frac12\ln|t|\\[6pt] &=\frac12\ln|x^2-2x-5| \end{aligned} \]
Hence,
\[\boxed{I_1=\frac12\ln|x^2-2x-5|}\]
Evaluate the second integral
\[I_2=4\int\frac{dx}{x^2-2x-5}\]
Complete the square in the denominator:
\[ \begin{aligned} x^2-2x-5 &=x^2-2x+1-1-5\\[5pt] &=(x-1)^2-6\\[5pt] &=(x-1)^2-(\sqrt6)^2 \end{aligned} \]
Therefore,
\[I_2=4\int\frac{dx}{(x-1)^2-(\sqrt6)^2}\]
Recall the standard formula:
\[ \int\frac{du}{u^2-a^2} =\frac{1}{2a}\ln\left|\frac{u-a}{u+a}\right|+C \]
Let
\[u=x-1,\quad du=dx,\quad a=\sqrt6.\]
Applying the formula,
\[ \begin{aligned} I_2 &=4\cdot\frac{1}{2\sqrt6} \ln\left|\frac{(x-1)-\sqrt6}{(x-1)+\sqrt6}\right|\\[6pt] &=\frac{2}{\sqrt6} \ln\left|\frac{x-1-\sqrt6}{x-1+\sqrt6}\right|\\[6pt] &=\frac{\sqrt6}{3} \ln\left|\frac{x-1-\sqrt6}{x-1+\sqrt6}\right| \end{aligned} \]
Thus,
\[\boxed{I_2=\frac{\sqrt6}{3}\ln\left|\frac{x-1-\sqrt6}{x-1+\sqrt6}\right|}\]
Combine both results
Since \(I=I_1+I_2\),
\[ \begin{aligned} I={}&\frac12\ln|x^2-2x-5|\\[5pt] &+\frac{\sqrt6}{3} \ln\left|\frac{x-1-\sqrt6}{x-1+\sqrt6}\right|+C \end{aligned} \]

💡 Answer

Final Answer

›

\[ \boxed{ \begin{aligned} \int\frac{x+3}{x^2-2x-5}\,dx ={}&\frac12\ln|x^2-2x-5|\\[5pt] &+\frac{\sqrt6}{3} \ln\left|\frac{x-1-\sqrt6}{x-1+\sqrt6}\right|+C \end{aligned} } \]

🎯 Exam Significance

Exam Significance

›

This problem tests decomposition of a linear numerator using the derivative of a quadratic denominator.
It combines logarithmic integration with the standard integral involving \(u^2-a^2\).
Correct coefficient comparison is essential. Here, the constant is \(B=4\), not \(B=1\).
Retaining absolute values in logarithmic expressions helps ensure the answer is valid on the appropriate intervals.

Significance for Competitive Entrance Examinations

The numerator-decomposition method is useful for rational functions whose numerator has degree one and whose denominator is quadratic.
Completing the square reveals the difference-of-squares structure needed for the second integral.
Careful tracking of constants, especially the outside factor \(4\), prevents common coefficient errors.
Differentiation is an efficient way to verify logarithmic antiderivatives.

🚫 Common Mistakes

Common Mistakes

›

Using \(B=1\). The correct equation is \(-2A+B=3\), so \(B=4\).
Changing the denominator from \(x^2-2x-5\) to \(x^2-2x+5\).
Forgetting to multiply the second integral by \(4\).
Using the wrong sign in the formula for \(\int\frac{du}{u^2-a^2}\).
Omitting absolute-value signs in logarithmic expressions.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

Express the numerator as a multiple of the derivative of the denominator plus a constant.
Use substitution to evaluate the logarithmic part.
Complete the square to identify the standard difference-of-squares integral.
Track every numerical coefficient carefully when splitting and combining integrals.
Verify the final expression by differentiation.

📘 Concept & Theory

Theory and Concept

›

This integral can be evaluated by expressing the numerator as a linear combination of the derivative of the quadratic expression under the square root and a constant. This allows us to split the original integral into two standard integrals.

The main concepts used are:

Decomposition of the numerator: Express the numerator in terms of the derivative of the quadratic expression.
Integration by substitution: Use the derivative of the quadratic expression to evaluate the first integral.
Completing the square: Rewrite the quadratic expression in the form \(u^2+a^2\).
Standard logarithmic integral:

\[ \int\frac{du}{\sqrt{u^2+a^2}} =\ln\left|u+\sqrt{u^2+a^2}\right|+C \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Express \(5x+3\) in terms of \(2x+4\) and a constant.
Determine the constants by comparing coefficients.
Split the original integral into two simpler integrals.
Evaluate the first integral using substitution.
Complete the square and evaluate the second integral using the standard logarithmic formula.
Combine the results and verify the answer by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 21 steps

Express the numerator appropriately.
Let
\[5x+3=A\frac{d}{dx}(x^2+4x+10)+B\]
Since
\[\frac{d}{dx}(x^2+4x+10)=2x+4,\]
we obtain
\[5x+3=A(2x+4)+B\]

\[5x+3=2Ax+4A+B\]
Comparing the coefficients of \(x\),
\[2A=5\]

\[A=\frac52\]
Comparing the constant terms,
\[4A+B=3\]
Substituting \(A=\frac52\),
\[4\left(\frac52\right)+B=3\]

\[10+B=3\]

\[B=-7\]
Therefore, the correct decomposition is
\[\boxed{5x+3=\frac52(2x+4)-7}\]
Split the integral
Substituting the decomposition into the original integral,
\[ \begin{aligned} I &=\int\frac{5x+3}{\sqrt{x^2+4x+10}}\,dx\\[6pt] &=\int\frac{\frac52(2x+4)-7} {\sqrt{x^2+4x+10}}\,dx\\[6pt] &=\frac52\int\frac{2x+4}{\sqrt{x^2+4x+10}}\,dx\\[6pt] &\quad-7\int\frac{dx}{\sqrt{x^2+4x+10}} \end{aligned} \]
Let
\[I_1=\frac52\int\frac{2x+4}{\sqrt{x^2+4x+10}}\,dx\]
and
\[I_2=-7\int\frac{dx}{\sqrt{x^2+4x+10}}.\]
Evaluate the first integral
\[I_1=\frac52\int\frac{2x+4}{\sqrt{x^2+4x+10}}\,dx\]
Use the substitution
\[t=x^2+4x+10\]
Differentiating with respect to \(x\),
\[dt=(2x+4)\,dx\]
Therefore,
\[ \begin{aligned} I_1 &=\frac52\int\frac{dt}{\sqrt t}\\[6pt] &=\frac52\int t^{-1/2}\,dt\\[6pt] &=\frac52\left(\frac{t^{1/2}}{1/2}\right)\\[6pt] &=5\sqrt t \end{aligned} \]
Substituting \(t=x^2+4x+10\), we get
\[\boxed{I_1=5\sqrt{x^2+4x+10}}\]
Evaluate the second integral
\[I_2=-7\int\frac{dx}{\sqrt{x^2+4x+10}}\]
Complete the square in the denominator:
\[ \begin{aligned} x^2+4x+10 &=x^2+4x+4+6\\[5pt] &=(x+2)^2+6\\[5pt] &=(x+2)^2+(\sqrt6)^2 \end{aligned} \]
Thus,
\[I_2=-7\int\frac{dx}{\sqrt{(x+2)^2+(\sqrt6)^2}}\]
Recall the standard integral formula:
\[ \int\frac{du}{\sqrt{u^2+a^2}} =\ln\left|u+\sqrt{u^2+a^2}\right|+C \]
Let
\[u=x+2,\quad du=dx,\quad a=\sqrt6.\]
Applying the formula,
\[ \begin{aligned} I_2 &=-7\ln\left|u+\sqrt{u^2+6}\right|+C\\[5pt] &=-7\ln\left|x+2+\sqrt{(x+2)^2+6}\right|+C \end{aligned} \]
Since \((x+2)^2+6=x^2+4x+10\),
\[ \boxed{ I_2=-7\ln\left|x+2+\sqrt{x^2+4x+10}\right| } \]
In fact, \(x+2+\sqrt{x^2+4x+10}>0\) for all real \(x\), so the absolute value around the logarithm's argument is optional.
Combine both results.
Since \(I=I_1+I_2\),
\[ \begin{aligned} I={}&5\sqrt{x^2+4x+10}\\ &-7\ln\left|x+2+\sqrt{x^2+4x+10}\right|+C \end{aligned} \]

💡 Answer

Final Answer

›

\[ \boxed{ \begin{aligned} \int\frac{5x+3}{\sqrt{x^2+4x+10}}\,dx ={}&5\sqrt{x^2+4x+10}\\ &-7\ln\left|x+2+\sqrt{x^2+4x+10}\right|+C \end{aligned} } \]

🎯 Exam Significance

Exam Significance

›

This question tests decomposition of a linear numerator using the derivative of a quadratic expression.
It combines substitution with a standard logarithmic integral.
Showing the coefficient comparison and completing-square steps clearly supports method marks.
Careful handling of the constant term is essential: the correct value is \(B=-7\).

Significance for Competitive Entrance Examinations

The decomposition method is useful for integrals involving a linear numerator and a quadratic expression under a square root.
Recognising the derivative of the quadratic expression simplifies one part of the integral.
Completing the square identifies the standard logarithmic form for the remaining integral.
Differentiation provides an efficient check of the final expression and its coefficients.

🚫 Common Mistakes

Common Mistakes

›

Calculating \(B\) incorrectly. From \(4A+B=3\) and \(A=\frac52\), the correct value is \(B=-7\).
Writing \(x^2+4x+10\) incorrectly when completing the square. It equals \((x+2)^2+6\).
Forgetting the coefficient \(-7\) when evaluating the second integral.
Omitting the constant of integration \(C\).
Not checking the final result by differentiation.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 5 points

Express the numerator in terms of the derivative of the quadratic expression and a constant.
Use substitution for the derivative-containing part.
Complete the square to obtain the standard form \(\int\frac{du}{\sqrt{u^2+a^2}}\).
Keep track of numerical coefficients while splitting and recombining the integrals.
The final antiderivative is valid for every real \(x\), because the quadratic expression under the square root is always positive.

📘 Concept & Theory

Theory and Concept

›

This question uses the method of completing the square and the standard inverse-tangent integral formula:

\[ \int\frac{du}{u^2+a^2} =\frac{1}{a}\tan^{-1}\left(\frac{u}{a}\right)+C \]

✏️ Solution

Complete Solution

›

Step-by-step Solution · 4 steps

Complete the square
\[ \begin{aligned} x^2+2x+2 &=x^2+2x+1+1\\ &=(x+1)^2+1 \end{aligned} \]
Therefore,
\[I=\int\frac{dx}{(x+1)^2+1}\]
Apply the standard formula
Let \(u=x+1\). Then \(du=dx\). Hence,
\[ \begin{aligned} I &=\int\frac{du}{u^2+1}\\[5pt] &=\tan^{-1}u+C\\[5pt] &=\tan^{-1}(x+1)+C \end{aligned} \]

💡 Answer

Final Answer

›

Correct Answer
\[\boxed{\text{Option (B): }\tan^{-1}(x+1)+C}\]

🎯 Exam Significance

Exam Significance

›

Recognise quadratic expressions that can be written as \((x+a)^2+b^2\).
Use substitution to match the denominator with the standard inverse-tangent formula.
In multiple-choice questions, differentiating an option can quickly verify whether it is correct.

🔑 Key Takeaways

Key Takeaways

›

Whenever the denominator can be expressed as \((x+a)^2+1\), the integral takes the form \(\tan^{-1}(x+a)+C\).

Q25

LONG ANSWER4 marks

Choose the correct answer \[\int \dfrac{dx}{\sqrt{9x-4x^{2}}}\] \[ \begin{aligned}\left( A\right) \quad &\dfrac{1}{9}\sin ^{-1}\left( \dfrac{9x-8}{8}\right) +c\\[6pt] \left( B\right) \quad &\dfrac{1}{2}\sin ^{-1}\left( \dfrac{8x-9}{9}\right) +C\\[6pt] \left( C\right) \quad &\dfrac{1}{3}\sin ^{-1}\left( \dfrac{9x-8}{s}\right) +c\\[6pt] \left( D\right) \quad &\dfrac{1}{2}\sin ^{-1}\left( \dfrac{9x-8}{9}\right) +C\end{aligned}\]

📘 Concept & Theory

Theory and Concept

›

This integral can be evaluated by completing the square in the quadratic expression under the square root and then applying the standard inverse-sine formula:

\[ \int\frac{du}{\sqrt{a^2-u^2}} =\sin^{-1}\left(\frac{u}{a}\right)+C \]

✏️ Solution

Complete Solution

›

Step-by-step Solution · 6 steps

Complete the square
\[ \begin{aligned} 9x-4x^2 &=-4\left(x^2-\frac94x\right)\\[5pt] &=-4\left[\left(x-\frac98\right)^2-\frac{81}{64}\right]\\[5pt] &=\frac{81}{16}-4\left(x-\frac98\right)^2\\[5pt] &=\frac{81-(8x-9)^2}{16} \end{aligned} \]
Therefore,
\[\sqrt{9x-4x^2}=\frac14\sqrt{81-(8x-9)^2}\]
The integral becomes
\[I=4\int\frac{dx}{\sqrt{81-(8x-9)^2}}\]
Apply substitution
Let
\[u=8x-9\]
Then
\[du=8\,dx\implies dx=\frac{du}{8}\]
Substituting,
\[ \begin{aligned} I &=4\int\frac{\frac18\,du}{\sqrt{81-u^2}}\\[5pt] &=\frac12\int\frac{du}{\sqrt{81-u^2}}\\[5pt] &=\frac12\sin^{-1}\left(\frac{u}{9}\right)+C \end{aligned} \]
Substituting \(u=8x-9\), we obtain
\[ \boxed{ I=\frac12\sin^{-1}\left(\frac{8x-9}{9}\right)+C } \]

💡 Answer

Final Answer

›

Correct Answer Option (B)

\[ \boxed{ \frac12\sin^{-1}\left(\frac{8x-9}{9}\right)+C } \]

🎯 Exam Significance

Exam Significance

›

Completing the square converts a quadratic expression into the standard inverse-sine form.
Substitution determines the coefficient in front of the inverse-sine function.
In multiple-choice questions, checking the argument and coefficient carefully helps eliminate incorrect options.
Differentiation is a reliable method for verifying an indefinite integral.

🔑 Key Takeaways

Key Takeaways

›

When integrating a reciprocal square root of a quadratic expression, complete the square first and then match the expression with the standard formula for \(\sin^{-1}\).

Integrals

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Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

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Integrals

6. Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

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Chapter Complete!

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Frequently Asked Questions

Q 01 What is covered in NCERT Class 12 Mathematics Chapter 7 Integrals Exercise 7.4?

Q 02 How can I solve integrals involving quadratic expressions in the denominator?

Q 03 Which standard formula is used to integrate 1/sqrt(a² - x²)?

Q 04 How do I integrate a function containing the square root of a quadratic expression?

Q 05 How can I integrate 1/(x² + a²)?

Q 06 How can I solve integrals involving 1/v(x² + a²)?

Q 07 Why is completing the square important in integration?

Q 08 How can I verify an indefinite integral answer?

Q 09 Are NCERT Class 12 Integrals solutions useful for CBSE board examinations?

Q 10 How can I prepare for competitive examinations using Integrals Exercise 7.4?

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