NCERT Class 12 Maths Integrals Ex 7.6 Solution

📘 Concept & Theory

Theory and Concept

›

Integration by parts is an important technique of integration used when the integrand is the product of two functions. It is derived from the product rule of differentiation and is particularly useful when direct integration is difficult.

The integration by parts formula is:

\[ \boxed{\int u\,v\,dx=u\int v\,dx-\int\left[\frac{du}{dx}\int v\,dx\right]dx} \]

Alternatively, the formula can be written in differential notation as:

\[ \boxed{\int u\,dv=uv-\int v\,du} \]

Here, one function is selected as \(u\), while the remaining part is taken as \(dv\). The choice of \(u\) is important because its differentiation should simplify the integrand.

🗺️ Solution Roadmap

Step-by-step Plan

›

Identify the two functions in the integrand \(x\sin x\).
Choose \(u=x\) and \(dv=\sin x\,dx\).
Differentiate \(u\) to obtain \(du=dx\).
Integrate \(dv\) to obtain \(v=-\cos x\).
Substitute these values into the integration by parts formula.
Simplify the resulting integral and evaluate it.
Add the constant of integration \(C\).

✏️ Solution

Complete Solution

›

Step-by-step Solution · 7 steps

Given —
\[I=\int x\sin x\,dx\]
Comparing the integrand with the integration by parts formula, choose:
\[u=x,\qquad dv=\sin x\,dx\]
Differentiate \(u\)
Differentiating \(u=x\) with respect to \(x\), we get:
\[du=\frac{d}{dx}(x)\,dx\]
\[\boxed{du=dx}\]
Integrate \(dv\).
Integrating \(dv=\sin x\,dx\), we obtain:
\[v=\int\sin x\,dx\]
\[\boxed{v=-\cos x}\]
Thus, the required values are:
\[u=x,\quad du=dx,\quad v=-\cos x\]
Apply the integration by parts formula
Using
\[\int u\,dv=uv-\int v\,du\]
Substituting \(u=x\), \(v=-\cos x\), and \(du=dx\), we get:
\[ \begin{aligned} I&=x(-\cos x)-\int(-\cos x)\,dx\\ &=-x\cos x+\int\cos x\,dx \end{aligned} \]
Evaluate the remaining integral
We know that:
\[\int\cos x\,dx=\sin x\]
Therefore:
\[I=-x\cos x+\sin x+C\]

💡 Answer

Final Answer

›

\[ \boxed{\displaystyle\int x\sin x\,dx =-x\cos x+\sin x+C} \]

🎯 Exam Significance

Exam Significance

›

Fundamental integration technique: This problem tests the application of integration by parts, an important technique in NCERT Class 12 Mathematics.
Formula application: Students must correctly identify \(u\), \(dv\), \(du\), and \(v\).
Product rule connection: It reinforces the relationship between differentiation and integration.
Step-wise marking: Correct selection of functions, substitution into the formula, simplification, and the final answer are important steps for scoring full marks.
Constant of integration: The arbitrary constant \(C\) must be included in an indefinite integral.

Significance for Competitive Entrance Examinations

JEE Main and Advanced: Integration by parts frequently appears in questions involving products of algebraic, trigonometric, exponential, and logarithmic functions.
Integration strategy: The problem develops the ability to select the function whose differentiation simplifies the integrand.
Inverse verification: Differentiating the final expression is an effective way to verify an indefinite integral and eliminate calculation errors.
Higher-level applications: The same technique is useful in evaluating definite integrals, reduction formulae, and more complicated integrals.

🚫 Common Mistakes

Common Mistakes

›

Forgetting the negative sign while integrating \(\sin x\).
Writing \(uv+\int v\,du\) instead of \(uv-\int v\,du\).
Making a sign error when substituting \(v=-\cos x\).
Omitting the constant of integration \(C\).
Skipping intermediate steps, particularly when applying the integration by parts formula.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

Integration by parts is derived from the product rule of differentiation.
The appropriate formula is \(\int u\,dv=uv-\int v\,du\).
For \(\int x\sin x\,dx\), choose \(u=x\) and \(dv=\sin x\,dx\).
The differential and integral of the selected functions are \(du=dx\) and \(v=-\cos x\), respectively.
The negative sign in \(v=-\cos x\) must be handled carefully while substituting into the formula.
Always include the arbitrary constant \(C\) in indefinite integration.
Differentiation provides a reliable method of checking the correctness of an indefinite integral.

📘 Concept & Theory

Theory and Concept

›

Integration by parts is used to integrate the product of two functions. It is particularly useful when differentiating one function simplifies it, while the other function can be integrated easily.

The integration by parts formula is:

\[ \boxed{\int u\,dv=uv-\int v\,du} \]

For an integral of the form \(\int f(x)g(x)\,dx\), we generally select one function as \(u\) and the remaining function as \(dv\). The selection should make the resulting integral simpler than the original one.

Important trigonometric integration formula:

\[ \int\sin ax\,dx=-\frac{\cos ax}{a}+C,\quad a\ne 0 \]

In this question, the presence of \(3x\) in the trigonometric function requires the application of the chain rule in reverse while integrating.

🗺️ Solution Roadmap

Step-by-step Plan

›

Identify the algebraic and trigonometric functions in the integrand.
Choose \(u=x\) and \(dv=\sin 3x\,dx\).
Differentiate \(u\) to obtain \(du=dx\).
Integrate \(dv\), taking care of the coefficient 3 inside the sine function.
Apply the integration by parts formula.
Evaluate the remaining trigonometric integral.
Simplify the expression and include the constant of integration.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 11 steps

Given —
\[I=\int x\sin 3x\,dx\]
Comparing with the integration by parts formula, choose:
\[u=x,\qquad dv=\sin 3x\,dx\]
Differentiate \(u\).
Differentiating \(u=x\) with respect to \(x\), we get:
\[du=\frac{d}{dx}(x)\,dx\]
\[\boxed{du=dx}\]
Integrate \(dv\).
We have:
\[v=\int\sin 3x\,dx\]
Using the standard formula:
\[\int\sin ax\,dx=-\frac{\cos ax}{a}\]
Putting \(a=3\), we obtain:
\[v=-\frac{\cos 3x}{3}\]
Thus, the required values are:
\[u=x,\quad du=dx,\quad v=-\frac{\cos 3x}{3}\]
Apply the integration by parts formula
Recall that:
\[\int u\,dv=uv-\int v\,du\]
Substituting the values of \(u\), \(v\), and \(du\):
\[ \begin{aligned} I&=uv-\int v\,du\\ &=x\left(-\frac{\cos 3x}{3}\right) -\int\left(-\frac{\cos 3x}{3}\right)dx\\ &=-\frac{x\cos 3x}{3} +\frac{1}{3}\int\cos 3x\,dx \end{aligned} \]
Evaluate the remaining integral
Using the standard trigonometric integration formula:
\[\int\cos ax\,dx=\frac{\sin ax}{a}+C\]
Putting \(a=3\), we get:
\[\int\cos 3x\,dx=\frac{\sin 3x}{3}\]
Substituting this result:
\[ \begin{aligned} I&=-\frac{x\cos 3x}{3} +\frac{1}{3}\left(\frac{\sin 3x}{3}\right)\\ &=-\frac{x\cos 3x}{3} +\frac{\sin 3x}{9} \end{aligned} \]
Adding the arbitrary constant of integration \(C\), we obtain:
\[\boxed{\displaystyle\int x\sin 3x\,dx=-\frac{x\cos 3x}{3}+\frac{\sin 3x}{9}+C}\]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int x\sin 3x\,dx=-\frac{x\cos 3x}{3}+\frac{\sin 3x}{9}+C}\]

🎯 Exam Significance

Exam Significance

›

Application of integration by parts: This question tests the ability to integrate the product of an algebraic function and a trigonometric function.
Trigonometric integration: Students must correctly apply the standard formula for integrating \(\sin ax\) and \(\cos ax\).
Chain rule connection: The coefficient 3 inside the trigonometric function must be accounted for during integration.
Step-wise presentation: Clearly showing the selection of \(u\), \(dv\), \(du\), and \(v\) helps demonstrate the method and secure method marks.
Algebraic accuracy: Correct handling of negative signs and fractional coefficients is essential for obtaining the final answer.

Significance for Competitive Entrance Examinations

JEE Main and Advanced: This type of integral develops proficiency in integration by parts, a frequently useful technique in higher-level integration problems.
Coefficient handling: Students learn to manage composite trigonometric functions involving expressions such as \(3x\), \(5x\), or \(ax\).
Conceptual understanding: It reinforces the relationship between integration by parts and the product rule of differentiation.
Advanced applications: The same approach can be extended to integrals involving \(x\cos ax\), \(x^2\sin ax\), and products of polynomial and exponential functions.
Error reduction: Verification by differentiation helps identify mistakes involving coefficients and signs.

🚫 Common Mistakes

Common Mistakes

›

Incorrect trigonometric integration: Writing \(\int\sin 3x\,dx=-\cos 3x\) instead of \(-\frac{\cos 3x}{3}\).
Incorrect argument: Replacing \(\cos 3x\) with \(\cos x\) while integrating.
Sign error: Forgetting that the integration by parts formula contains a subtraction sign.
Incorrect coefficient: Writing \(\frac{\sin 3x}{3}\) instead of \(\frac{\sin 3x}{9}\) in the final expression.
Omitting the constant: Forgetting to add \(C\) in an indefinite integral.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 8 points

Use integration by parts when the integrand is a product of two functions.
The formula is \(\int u\,dv=uv-\int v\,du\).
For this problem, the appropriate selection is \(u=x\) and \(dv=\sin 3x\,dx\).
When integrating \(\sin 3x\), the coefficient 3 appears in the denominator.
Remember that \(\int\sin 3x\,dx=-\frac{\cos 3x}{3}\).
The negative sign in the integration by parts formula must be handled carefully.
Always include the arbitrary constant \(C\) in an indefinite integral.
Differentiation is an effective method for verifying the final result.

📘 Concept & Theory

Theory and Concept

›

Integration by parts is a technique used to evaluate integrals involving the product of two functions. It is particularly useful when one function becomes simpler after successive differentiation, while the other function can be integrated repeatedly without becoming more complicated.

The integration by parts formula is:

\[ \boxed{\int u\,dv=uv-\int v\,du} \]

For an integral involving a polynomial multiplied by an exponential function, repeated integration by parts is an effective approach.

In this problem, the algebraic function \(x^2\) becomes simpler through successive differentiation:

\[ x^2\longrightarrow 2x\longrightarrow 2\longrightarrow 0 \]

On the other hand, repeated integration of \(e^x\) gives the same function:

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

Therefore, integration by parts must be applied twice to evaluate the given integral completely.

🗺️ Solution Roadmap

Step-by-step Plan

›

Identify the polynomial function \(x^2\) and exponential function \(e^x\).
Choose \(u=x^2\)` and \(dv=e^x\,dx\).
Differentiate \(u\) and integrate \(dv\).
Apply integration by parts for the first time.
Evaluate the remaining integral \(\int 2xe^x\,dx\) using integration by parts again.
Simplify the resulting expression carefully, keeping track of negative signs and coefficients.
Factor out \(e^x\) and add the constant of integration.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 17 steps

Given —
\[\int x^2e^x\,dx\]
Choose:
\[u=x^2,\qquad dv=e^x\,dx\]
Differentiate \(u\) and integrate \(dv\).
Differentiating \(u=x^2\), we obtain:
\[du=\frac{d}{dx}(x^2)\,dx\]
\[\boxed{du=2x\,dx}\]
Integrating \(dv=e^x\,dx\), we get:
\[v=\int e^x\,dx\]

\[\boxed{v=e^x}\]
Thus:
\[u=x^2,\quad du=2x\,dx,\quad v=e^x\]
Apply integration by parts for the first time
Using the formula:
\[\int u\,dv=uv-\int v\,du\]
Substituting the values:
\[ \begin{aligned} I&=x^2e^x-\int e^x(2x)\,dx\\ &=x^2e^x-2\int xe^x\,dx \end{aligned} \]
We now need to evaluate the remaining integral:
\[J=\int xe^x\,dx\]
Apply integration by parts for the second time.
For the integral:
\[J=\int xe^x\,dx\]
Choose:
\[u=x,\quad dv=e^x\,dx\]
Differentiating \(u=x\):
\[du=dx\]
Integrating \(dv=e^x\,dx\):
\[v=e^x\]
Applying integration by parts:
\[ \begin{aligned} J&=uv-\int v\,du\\ &=xe^x-\int e^x\,dx\\ &=xe^x-e^x \end{aligned} \]
Therefore:
\[\boxed{J=e^x(x-1)}\]
Substitute the value of \(J\) into the original integral.
We obtained:
\[I=x^2e^x-2J\]
Substituting \(J=xe^x-e^x\):
\[ \begin{aligned} I&=x^2e^x-2(xe^x-e^x)\\ &=x^2e^x-2xe^x+2e^x \end{aligned} \]
Taking \(e^x\) common from all the terms:
\[I=e^x(x^2-2x+2)\]
Adding the arbitrary constant of integration \(C\), we obtain:
\[\boxed{\displaystyle\int x^2e^x\,dx=e^x(x^2-2x+2)+C}\]

💡 Answer

Final Answer

›

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

🎯 Exam Significance

Exam Significance

›

Repeated integration by parts: This problem demonstrates how integration by parts can be applied more than once to evaluate an integral.
Polynomial differentiation: Students must correctly differentiate the polynomial successively and simplify the resulting expressions.
Exponential integration: It reinforces the important property that the derivative and indefinite integral of \(e^x\) are both \(e^x\), apart from the constant of integration.
Algebraic accuracy: Correct handling of brackets, coefficients, and negative signs is essential for obtaining full marks.
Presentation of working: Showing both applications of integration by parts makes the mathematical reasoning clear and supports step-wise marking.

Significance for Competitive Entrance Examinations

JEE Main and Advanced: Polynomial-exponential integrals are useful models for developing techniques required in more complex integration problems.
Repeated integration: Students learn to recognize when a second or subsequent application of integration by parts is necessary.
Tabular integration: The problem introduces an efficient computational strategy for integrals involving polynomials multiplied by exponential or trigonometric functions.
Reduction in calculation time: Recognizing the pattern of successive differentiation and integration can simplify lengthy calculations.
Higher-level applications: The same technique is useful in reduction formulae, definite integrals, and advanced mathematical applications.

🚫 Common Mistakes

Common Mistakes

›

Stopping too early: After the first application of integration by parts, the integral \(\int xe^x\,dx\) still needs to be evaluated.
Incorrect differentiation: Remember that \(\frac{d}{dx}(x^2)=2x\), not \(x\).
Coefficient error: The factor 2 obtained during differentiation must be retained throughout the calculation.
Sign error: Carefully apply the subtraction in the integration by parts formula.
Bracket error: When substituting \(xe^x-e^x\), distribute the factor 2 correctly.
Missing constant: Include \(C\) in the final indefinite integral.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 9 points

Integration by parts is based on the product rule of differentiation.
The fundamental formula is \(\int u\,dv=uv-\int v\,du\).
For polynomial-exponential integrals, selecting the polynomial as \(u\) is generally effective.
When integrating \(e^x\), the result remains \(e^x\).
Repeated integration by parts is necessary when the remaining integral still contains a product of two functions.
Successive differentiation of \(x^2\) eventually gives zero, which makes this integral suitable for the tabular method.
The tabular method provides a compact alternative to writing repeated integration by parts separately.
Always verify indefinite integrals by differentiating the final expression.
Do not forget the arbitrary constant \(C\).

📘 Concept & Theory

Theory and Concept

›

Integration by parts is used to evaluate integrals involving the product of two functions. It is particularly useful when one function becomes simpler after differentiation and the other can be integrated directly.

The integration by parts formula is:

\[ \boxed{\int u\,dv=uv-\int v\,du} \]

ILATE Rule for Selecting \(u\)

ILATE is a commonly used guideline for selecting the function to be differentiated (that is, \(u\)) when applying integration by parts. According to this rule, preference is generally given in the following order:

I: Inverse trigonometric functions
L: Logarithmic functions
A: Algebraic functions
T: Trigonometric functions
E: Exponential functions

Since the logarithmic function \(\log x\) has higher priority than the algebraic function \(x\), we select \(\log x\) as \(u\) and \(x\,dx\) as \(dv\).

Important formulas used:

\[ \frac{d}{dx}(\log x)=\frac{1}{x},\quad x>0 \]

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

🗺️ Solution Roadmap

Step-by-step Plan

›

Identify the logarithmic and algebraic functions in the integrand.
Apply the ILATE rule to select \(u=\log x\).
Take \(dv=x\,dx\) and calculate \(du\) and \(v\).
Substitute these values into the integration by parts formula.
Simplify the remaining integral by cancelling the common factor \(x\).
Integrate the resulting algebraic expression.
Factorise the result and include the constant of integration.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 12 steps

Given —
\[I=\int x\log x\,dx\]
Using the ILATE rule, the logarithmic function is selected first. Therefore, take:
\[u=\log x,\quad dv=x\,dx\]
Differentiate \(u\).
Differentiating \(u=\log x\) with respect to \(x\):
\[du=\frac{d}{dx}(\log x)\,dx\]
Using the standard derivative formula:
\[\boxed{du=\frac{1}{x}\,dx}\]
Integrate \(dv\)
We have:
\[v=\int x\,dx\]
Applying the power rule of integration:
\[v=\frac{x^{2}}{2}\]
Thus, the required values are:
\[u=\log x,\quad du=\frac{1}{x}\,dx,\quad v=\frac{x^2}{2}\]
Apply the integration by parts formula.
Recall that:
\[\int u\,dv=uv-\int v\,du\]
Substituting the values of \(u\), \(v\), and \(du\):
\[ \begin{aligned} I&=uv-\int v\,du\\ &=\log x\cdot\frac{x^2}{2} -\int\frac{x^2}{2}\cdot\frac{1}{x}\,dx\\ &=\frac{x^2}{2}\log x -\frac{1}{2}\int\frac{x^2}{x}\,dx \end{aligned} \]
Simplify the remaining integral
Since:
\[\frac{x^2}{x}=x\]
We obtain:
\[I=\frac{x^2}{2}\log x-\frac{1}{2}\int x\,dx\]
Evaluate the remaining integral
Using the power rule
\[\int x\,dx=\frac{x^2}{2}\]
Substituting:
\[ \begin{aligned} I&=\frac{x^2}{2}\log x -\frac{1}{2}\left(\frac{x^2}{2}\right)\\ &=\frac{x^2}{2}\log x-\frac{x^2}{4} \end{aligned} \]
Taking \(\frac{x^2}{2}\) common:
\[I=\frac{x^2}{2}\left(\log x-\frac{1}{2}\right)\]
Adding the arbitrary constant of integration \(C\):
\[\boxed{\displaystyle\int x\log x\,dx=\frac{x^2}{2}\left(\log x-\frac{1}{2}\right)+C}\]

💡 Answer

Final Answer

›

🎯 Exam Significance

Exam Significance

›

Application of ILATE: This problem demonstrates the selection of the logarithmic function as \(u\), which is an important application of the ILATE guideline.
Integration by parts: Students learn how to integrate the product of a logarithmic function and an algebraic function.
Derivative of logarithmic functions: It reinforces the standard result \(\frac{d}{dx}(\log x)=\frac{1}{x}\).
Power rule: The problem also tests the ability to integrate algebraic expressions accurately.
Step-wise marking: Correct selection of \(u\) and \(dv\), application of the formula, simplification, and the final result are important for securing method marks.

Significance for Competitive Entrance Examinations

JEE Main and Advanced: Logarithmic integrals are important in developing proficiency in integration by parts and evaluating more complex expressions.
ILATE-based selection: Understanding the priority of logarithmic and algebraic functions helps students make an efficient choice of \(u\).
Algebraic simplification: The problem develops accuracy in simplifying expressions before performing further integration.
Conceptual application: Similar techniques can be extended to integrals involving \(x^n\log x\), \(\log x\), and products of logarithmic and exponential functions.
Verification skills: Differentiation of the final expression helps identify errors in coefficients and algebraic manipulation.

🚫 Common Mistakes

Common Mistakes

›

Incorrect selection of \(u\): Choosing \(x\) instead of \(\log x\) ignores the useful ILATE-based selection for this problem.
Incorrect differentiation: Remember that the derivative of \(\log x\) is \(\frac{1}{x}\), not \(\log x\).
Coefficient error: Do not omit the factor \(\frac{1}{2}\) obtained when integrating \(x\).
Incorrect simplification: Carefully simplify \(\frac{x^2}{2}\cdot\frac{1}{x}\) to \(\frac{x}{2}\).
Sign error: Retain the subtraction sign in the integration by parts formula.
Domain consideration: The real-valued expression \(\log x\) requires \(x>0\).
Missing constant: Do not forget the arbitrary constant \(C\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 8 points

Integration by parts is particularly useful for integrating products of logarithmic and algebraic functions.
The integration by parts formula is \(\int u\,dv=uv-\int v\,du\).
According to ILATE, logarithmic functions are given preference over algebraic functions when selecting \(u\).
For this problem, select \(u=\log x\) and \(dv=x\,dx\).
Remember that \(\frac{d}{dx}(\log x)=\frac{1}{x}\).
Simplify the remaining integral before applying another integration technique.
Always include the arbitrary constant \(C\) in an indefinite integral.
Verify the final result by differentiation whenever possible.

📘 Concept & Theory

Theory and Concept

›

Integration by parts is used to evaluate integrals involving the product of two functions. It is particularly effective when differentiating one function simplifies it and the other function can be integrated easily.

The integration by parts formula is:

\[ \boxed{\int u\,dv=uv-\int v\,du} \]

ILATE Rule

ILATE provides a guideline for selecting the function to be differentiated (u) in integration by parts. The priority order is:

I: Inverse trigonometric functions
L: Logarithmic functions
A: Algebraic functions
T: Trigonometric functions
E: Exponential functions

In the given integral, \(\log 2x\) is a logarithmic function and \(x\) is an algebraic function. According to ILATE, we select the logarithmic function as \(u\).

Important derivative formula:

\[ \frac{d}{dx}(\log 2x) =\frac{1}{2x}\cdot 2 =\frac{1}{x} \]

The derivative is calculated using the chain rule. The factor 2 from differentiating the inner function cancels with the denominator 2.

🗺️ Solution Roadmap

Step-by-step Plan

›

Identify the logarithmic and algebraic functions.
Apply ILATE and choose \(u=\log 2x\) and \(dv=x\,dx\).
Differentiate \(u\) using the chain rule.
Integrate \(dv\) using the power rule.
Apply the integration by parts formula.
Simplify and evaluate the remaining algebraic integral.
Write the final result with the constant of integration.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 13 steps

Given —
\[I=\int x\log 2x\,dx\]
Using the ILATE rule, select:
\[u=\log 2x,\quad dv=x\,dx\]
Differentiating \(u=\log 2x\) with respect to \(x\), we get:
\[du=\frac{d}{dx}(\log 2x)\,dx\]
Applying the chain rule:
\[ \begin{aligned} du&=\frac{1}{2x}\cdot\frac{d}{dx}(2x)\,dx\\ &=\frac{1}{2x}\cdot 2\,dx\\ &=\frac{1}{x}\,dx \end{aligned} \]
Therefore:
\[\boxed{du=\frac{1}{x}\,dx}\]
Integrate \(dv\)
We have:
\[v=\int x\,dx\]
Using the power rule of integration:
\[\boxed{v=\frac{x^2}{2}}\]
Thus, the required values are:
\[ u=\log 2x,\quad du=\frac{1}{x}\,dx,\quad v=\frac{x^2}{2} \]
Apply the integration by parts formula.
Recall that:
\[\int u\,dv=uv-\int v\,du\]
Substituting the values of \(u\), \(v\), and \(du\):
\[ \begin{aligned} I&=uv-\int v\,du\\ &=\log 2x\cdot\frac{x^2}{2} -\int\frac{x^2}{2}\cdot\frac{1}{x}\,dx\\ &=\frac{x^2}{2}\log 2x -\frac{1}{2}\int\frac{x^2}{x}\,dx \end{aligned} \]
Simplify the remaining integral
Since:
\[\frac{x^2}{x}=x\]
We obtain:
\[I=\frac{x^2}{2}\log 2x-\frac{1}{2}\int x\,dx\]
Evaluate the remaining integral
Using the power rule:
\[\int x\,dx=\frac{x^2}{2}\]
Substituting this value:
\[ \begin{aligned} I&=\frac{x^2}{2}\log 2x -\frac{1}{2}\left(\frac{x^2}{2}\right)\\ &=\frac{x^2}{2}\log 2x-\frac{x^2}{4} \end{aligned} \]
Taking \(\frac{x^2}{2}\) common:
\[I=\frac{x^2}{2}\left(\log 2x-\frac{1}{2}\right)\]
Adding the arbitrary constant of integration \(C\):
\[\boxed{\displaystyle\int x\log 2x\,dx=\frac{x^2}{2}\left(\log 2x-\frac{1}{2}\right)+C}\]

💡 Answer

Final Answer

›

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

🎯 Exam Significance

Exam Significance

›

Application of ILATE: This problem tests the selection of the logarithmic function as \(u\) when applying integration by parts.
Chain rule: Students must correctly differentiate the composite logarithmic function \(\log 2x\).
Integration by parts: It reinforces the application of the standard formula to a logarithmic-algebraic product.
Algebraic simplification: The problem tests the ability to simplify fractional algebraic expressions without losing coefficients.
Step-wise presentation: Clearly showing the calculation of \(du\), \(v\), and the remaining integral helps demonstrate the complete mathematical procedure.

Significance for Competitive Entrance Examinations

JEE Main and Advanced: Logarithmic integrals involving linear expressions inside the logarithm are useful practice for integration by parts.
Chain rule application: The problem reinforces differentiation of composite functions and the cancellation of inner-function coefficients.
Alternative approaches: The logarithmic identity \(\log 2x=\log 2+\log x\) provides another way to evaluate the integral.
Conceptual flexibility: Students learn to recognize equivalent integral forms and choose an efficient method.
Accuracy under time constraints: Correct handling of logarithmic identities, coefficients, and signs is essential in objective and numerical-answer questions.

🚫 Common Mistakes

Common Mistakes

›

Incorrect differentiation: Writing \(\frac{d}{dx}(\log 2x)=\frac{1}{2x}\) without multiplying by the derivative of \(2x\).
Incorrect chain rule: Forgetting that \(\frac{d}{dx}(2x)=2\).
Coefficient error: Omitting the factor \(\frac{1}{2}\) obtained from integrating \(x\).
Incorrect simplification: Failing to simplify \(\frac{x^2}{2}\cdot\frac{1}{x}\) to \(\frac{x}{2}\).
Logarithmic notation: Confusing \(\log 2x\) with \(2\log x\). In general, \(\log 2x=\log 2+\log x\).
Missing constant: Forgetting to add \(C\) to the indefinite integral.
Domain restriction: For real-valued logarithms, \(2x>0\), so \(x>0\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 8 points

Integration by parts is useful for integrating products of logarithmic and algebraic functions.
The ILATE rule suggests choosing the logarithmic function as \(u\).
For this integral, select \(u=\log 2x\) and \(dv=x\,dx\).
The chain rule gives \(\frac{d}{dx}(\log 2x)=\frac{1}{x}\).
The integral of \(x\) is \(\frac{x^2}{2}\).
The logarithmic identity \(\log 2x=\log 2+\log x\) provides an alternative solution.
Always retain the constant of integration \(C\).
Differentiation can be used to verify the correctness of the final result.

📘 Concept & Theory

Theory and Concept

›

Integration by parts is a fundamental technique of integration used to evaluate products of two functions. It is particularly useful when differentiating one function simplifies it and integrating the other function is straightforward.

The integration by parts formula is:

\[ \boxed{\int u\,dv=uv-\int v\,du} \]

ILATE Rule for Selecting \(u\)

ILATE is a useful guideline for deciding which function should be differentiated while applying integration by parts. The priority order is:

I: Inverse trigonometric functions
L: Logarithmic functions
A: Algebraic functions
T: Trigonometric functions
E: Exponential functions

In the given integral, \(\log x\) is a logarithmic function and \(x^2\) is an algebraic function. According to ILATE, the logarithmic function is selected as \(u\), while the algebraic function is taken as \(dv\).

Important formulas used:

\[ \frac{d}{dx}(\log x)=\frac{1}{x},\quad x>0 \]

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

🗺️ Solution Roadmap

Step-by-step Plan

›

Identify the logarithmic and algebraic functions in the integrand.
Apply the ILATE rule and select \(u=\log x\) and \(dv=x^2\,dx\).
Differentiate the logarithmic function to obtain \(du\).
Integrate the algebraic function to obtain \(v\).
Substitute the values into the integration by parts formula.
Simplify the remaining integral using the laws of exponents.
Evaluate the resulting polynomial integral and include the constant of integration.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 14 steps

Given —
\[I=\int x^2\log x\,dx\]
Using the ILATE rule, choose:
\[u=\log x,\qquad dv=x^2\,dx\]
Differentiate \(u\).
Differentiating \(u=\log x\) with respect to \(x\):
\[du=\frac{d}{dx}(\log x)\,dx\]
Using the standard derivative formula:
\[\boxed{du=\frac{1}{x}\,dx}\]
Integrate \(dv\)
We have:
\[v=\int x^2\,dx\]
Applying the power rule of integration:
\[ \begin{aligned} v&=\frac{x^{2+1}}{2+1}\\ &=\frac{x^3}{3} \end{aligned} \]
Therefore:
\[\boxed{v=\frac{x^3}{3}}\]
The required values are:
\[u=\log x,\quad du=\frac{1}{x}\,dx,\quad v=\frac{x^3}{3}\]
Apply the integration by parts formula
Recall that:
\[\int u\,dv=uv-\int v\,du\]
Substituting the values of \(u\), \(v\), and \(du\):
\[ \begin{aligned} I&=uv-\int v\,du\\ &=\log x\cdot\frac{x^3}{3} -\int\frac{x^3}{3}\cdot\frac{1}{x}\,dx\\ &=\frac{x^3}{3}\log x -\frac{1}{3}\int\frac{x^3}{x}\,dx \end{aligned} \]
Simplify the remaining integral
Using the law of exponents:
\[\frac{x^3}{x}=x^{3-1}=x^2\]
Therefore:
\[I=\frac{x^3}{3}\log x-\frac{1}{3}\int x^2\,dx\]
Evaluate the remaining integral
Using the power rule:
\[\int x^2\,dx=\frac{x^3}{3}\]
Substituting this value:
\[\begin{aligned}I&=\frac{x^3}{3}\log x-\frac{1}{3}\left(\frac{x^3}{3}\right)\\ &=\frac{x^3}{3}\log x-\frac{x^3}{9}\end{aligned}\]
Taking \(\frac{x^3}{9}\) common:
\[I=\frac{x^3}{9}(3\log x-1)\]
Adding the arbitrary constant of integration \(C\), we obtain:
\[\boxed{\displaystyle\int x^2\log x\,dx=\frac{x^3}{3}\log x-\frac{x^3}{9}+C}\]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int x^2\log x\,dx=\frac{x^3}{3}\log x-\frac{x^3}{9}+C}\]

🎯 Exam Significance

Exam Significance

›

Application of ILATE: This problem tests the selection of the logarithmic function as \(u\) and the algebraic function as \(dv\).
Integration by parts: It reinforces the application of the standard formula to products involving logarithmic and polynomial functions.
Power rule: Students must correctly integrate \(x^2\) to obtain \(\frac{x^3}{3}\).
Algebraic simplification: The problem tests the application of exponent laws while simplifying the remaining integral.
Step-wise marking: Clearly presenting the selection of functions, differentiation, integration, substitution, and simplification helps demonstrate the complete method.
Presentation of final results: Factorisation and inclusion of the integration constant improve mathematical presentation.

Significance for Competitive Entrance Examinations

JEE Main and Advanced: Integrals involving polynomial and logarithmic functions are useful models for developing proficiency in integration by parts.
ILATE-based selection: The problem strengthens the ability to identify the most appropriate function for differentiation.
Generalisation: The same approach can be extended to integrals of the form \(\int x^n\log x\,dx\), where \(n\) is a constant and \(n\ne -1\).
Algebraic accuracy: Students develop precision in handling powers, fractional coefficients, and logarithmic expressions.
Higher-level applications: This technique is useful in evaluating more complicated integrals, definite integrals, and reduction formulae.
Time management: Recognising the ILATE pattern helps students select an efficient integration strategy in competitive examinations.

🚫 Common Mistakes

Common Mistakes

›

Incorrect selection of \(u\): Choosing \(x^2\) instead of \(\log x\) is generally less efficient for this integral.
Incorrect differentiation: Remember that \(\frac{d}{dx}(\log x)=\frac{1}{x}\).
Incorrect integration: The integral of \(x^2\) is \(\frac{x^3}{3}\), not \(\frac{x^2}{3}\).
Power simplification error: Remember that \(\frac{x^3}{x}=x^2\).
Coefficient error: Do not lose the factor \(\frac{1}{3}\) when evaluating the remaining integral.
Sign error: Maintain the negative sign in the integration by parts formula.
Missing constant: Always include \(C\) in an indefinite integral.
Domain restriction: For the real-valued logarithmic function, \(x>0\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 9 points

Integration by parts is particularly useful for products of logarithmic and algebraic functions.
The standard formula is \(\int u\,dv=uv-\int v\,du\).
According to ILATE, logarithmic functions receive priority over algebraic functions when selecting \(u\).
For this problem, choose \(u=\log x\) and \(dv=x^2\,dx\).
The derivative of \(\log x\) is \(\frac{1}{x}\).
The integral of \(x^2\) is \(\frac{x^3}{3}\).
Carefully simplify the remaining integral before evaluating it.
The method can be generalised to integrals involving \(x^n\log x\).
Always include the arbitrary constant \(C\) and verify the result through differentiation.

📘 Concept & Theory

Theory and Concept

›

This problem involves the product of an algebraic function \(x\) and an inverse trigonometric function \(\sin^{-1}x\). Integration by parts is the appropriate technique because differentiating the inverse trigonometric function simplifies the integrand.

The integration by parts formula is:

\[ \boxed{\int u\,dv=uv-\int v\,du} \]

ILATE Rule

ILATE provides a guideline for selecting the function to be differentiated. Its priority order is:

I: Inverse trigonometric functions
L: Logarithmic functions
A: Algebraic functions
T: Trigonometric functions
E: Exponential functions

Since \(\sin^{-1}x\) is an inverse trigonometric function and \(x\) is an algebraic function, ILATE suggests choosing \(\sin^{-1}x\) as \(u\).

Important derivative formula:

\[ \frac{d}{dx}(\sin^{-1}x) =\frac{1}{\sqrt{1-x^2}},\quad |x|<1 \]

After applying integration by parts, the problem reduces to evaluating an integral involving \(\frac{x^2}{\sqrt{1-x^2}}\). This requires algebraic rearrangement and the standard integral of \(\sqrt{1-x^2}\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Apply the ILATE rule and choose \(u=\sin^{-1}x\) and \(dv=x\,dx\).
Calculate \(du\) using the derivative of the inverse sine function.
Integrate \(dv\) to obtain \(v=\frac{x^2}{2}\).
Apply integration by parts to obtain a simpler expression.
Evaluate the remaining integral \(\int\frac{x^2}{\sqrt{1-x^2}}\,dx\) by rewriting \(x^2=1-(1-x^2)\).
Use the standard integral formula for \(\sqrt{1-x^2}\).
Substitute the result into the original expression and simplify carefully.
Verify the final answer by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 23 steps

Given —
\[I=\int x\sin^{-1}x\,dx\]
Using the ILATE rule, select:
\[u=\sin^{-1}x,\quad dv=x\,dx\]
Differentiate \(u\).
Differentiating \(u=\sin^{-1}x\), we obtain:
\[du=\frac{d}{dx}(\sin^{-1}x)\,dx\]
Using the standard derivative formula:
\[\boxed{du=\frac{1}{\sqrt{1-x^2}}\,dx}\]
Integrate \(dv\).
We have:
\[v=\int x\,dx\]
Applying the power rule:
\[\boxed{v=\frac{x^2}{2}}\]
Thus, the required values are:
\[u=\sin^{-1}x,\quad du=\frac{1}{\sqrt{1-x^2}}\,dx,\quad v=\frac{x^2}{2}\]
Apply integration by parts
Recall the formula:
\[\int u\,dv=uv-\int v\,du\]
Substituting the values:
\[ \begin{aligned} I&=uv-\int v\,du\\ &=\frac{x^2}{2}\sin^{-1}x -\int\frac{x^2}{2} \cdot\frac{1}{\sqrt{1-x^2}}\,dx\\ &=\frac{x^2}{2}\sin^{-1}x -\frac{1}{2}\int \frac{x^2}{\sqrt{1-x^2}}\,dx \end{aligned} \]
Let the remaining integral be:
\[J=\int\frac{x^2}{\sqrt{1-x^2}}\,dx\]
Simplify the remaining integral
We know that:
\[x^2=1-(1-x^2)\]
Substituting this identity into \(J\):
\[\begin{aligned}J&=\int\frac{1-(1-x^2)}{\sqrt{1-x^2}}\,dx\\ &=\int\left[\frac{1}{\sqrt{1-x^2}}-\frac{1-x^2}{\sqrt{1-x^2}}\right]dx\\ &=\int\frac{1}{\sqrt{1-x^2}}\,dx-\int\sqrt{1-x^2}\,dx\end{aligned}\]
Therefore:
\[\boxed{J=\sin^{-1}x-\int\sqrt{1-x^2}\,dx}\]
Evaluate the standard integral
We use the standard formula:
\[\boxed{\int\sqrt{a^2-x^2}\,dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)+C}\]
Comparing \(\sqrt{1-x^2}\) with \(\sqrt{a^2-x^2}\), we have:
\[a^2=1\quad\Rightarrow\quad a=1\]
Substituting \(a=1\) into the formula:
\[\begin{aligned}\int\sqrt{1-x^2}\,dx&=\frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}\sin^{-1}x\\ &=\frac{1}{2}\left[x\sqrt{1-x^2}+\sin^{-1}x\right]\end{aligned}\]
Hence:
\[\boxed{\int\sqrt{1-x^2}\,dx=\frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}\sin^{-1}x+C}\]
Substitute the standard integral into \(J\)
Recall that:
\[J=\sin^{-1}x-\int\sqrt{1-x^2}\,dx\]
Substituting the evaluated integral:
\[\begin{aligned}J&=\sin^{-1}x-\left[\frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}\sin^{-1}x\right]\\ &=\sin^{-1}x-\frac{x}{2}\sqrt{1-x^2}-\frac{1}{2}\sin^{-1}x\\ &=\frac{1}{2}\sin^{-1}x-\frac{x}{2}\sqrt{1-x^2}\end{aligned}\]
Therefore:
\[\boxed{J=\frac{1}{2}\left[\sin^{-1}x-x\sqrt{1-x^2}\right]}\]
Substitute \(J\) into the original integral.
\[I=\frac{x^2}{2}\sin^{-1}x-\frac{1}{2}J\]
Substituting the value of \(J\):
\[\begin{aligned}I&=\frac{x^2}{2}\sin^{-1}x-\frac{1}{2}\left[\frac{1}{2}\sin^{-1}x-\frac{x}{2}\sqrt{1-x^2}\right]\\ &=\frac{x^2}{2}\sin^{-1}x-\frac{1}{4}\sin^{-1}x+\frac{x}{4}\sqrt{1-x^2}\end{aligned}\]
Taking \(\frac{1}{4}\) common:
\[I=\frac{1}{4}\left[(2x^2-1)\sin^{-1}x+x\sqrt{1-x^2}\right]\]
Adding the arbitrary constant of integration \(C\), we obtain:
\[\boxed{\displaystyle\int x\sin^{-1}x\,dx=\frac{1}{4}\left[(2x^2-1)\sin^{-1}x+x\sqrt{1-x^2}\right]+C}\]

💡 Answer

Final Answer

›

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

🎯 Exam Significance

Exam Significance

›

Application of ILATE: This problem tests the selection of an inverse trigonometric function as \(u\) while applying integration by parts.
Inverse trigonometric derivatives: Students must correctly use the derivative of \(\sin^{-1}x\).
Standard integral formula: It tests knowledge and application of the integral involving \(\sqrt{a^2-x^2}\).
Algebraic manipulation: The identity \(x^2=1-(1-x^2)\) is essential for simplifying the remaining integral.
Multi-step problem-solving: The question combines integration by parts, standard integration formulas, and algebraic simplification.
Step-wise marking: Clearly presenting each transformation helps demonstrate the complete mathematical procedure.

Significance for Competitive Entrance Examinations

JEE Main and Advanced: This problem develops proficiency in combining integration by parts with inverse trigonometric identities and standard integrals.
Conceptual integration: Students learn to recognise when an apparently complicated integral can be simplified using an algebraic identity.
Standard integral recognition: Familiarity with \(\int\sqrt{a^2-x^2}\,dx\) is useful in several advanced integration problems.
Analytical ability: The problem develops the ability to break a complicated integral into manageable parts.
Verification skills: Differentiating the final expression is particularly useful for checking complicated answers involving inverse trigonometric functions.

🚫 Common Mistakes

Common Mistakes

›

Incorrect selection: Choosing \(x\) instead of \(\sin^{-1}x\) as \(u\) ignores the ILATE priority.
Incorrect derivative: Remember that \(\frac{d}{dx}(\sin^{-1}x)=\frac{1}{\sqrt{1-x^2}}\).
Algebraic identity error: Use \(x^2=1-(1-x^2)\), not \(x^2=(1-x^2)-1\).
Sign error: Carefully distinguish between \(\int\frac{x^2}{\sqrt{1-x^2}}\,dx\) and its equivalent simplified expression.
Incorrect standard formula: The term \(\frac{x}{2}\sqrt{1-x^2}\) must contain the factor \(x\).
Substitution error: Remember that the remaining integral is multiplied by \(-\frac{1}{2}\) in the original expression.
Missing constant: Include \(C\) in the final indefinite integral.
Domain restriction: The real-valued derivative formula applies for \(-1

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 9 points

Integration by parts is effective for products involving inverse trigonometric and algebraic functions.
According to ILATE, inverse trigonometric functions receive the highest priority when selecting \(u\).
For this problem, choose \(u=\sin^{-1}x\) and \(dv=x\,dx\).
The derivative of \(\sin^{-1}x\) is \(\frac{1}{\sqrt{1-x^2}}\).
The identity \(x^2=1-(1-x^2)\) simplifies the remaining integral.
The standard integral of \(\sqrt{a^2-x^2}\) is important for evaluating the resulting expression.
Careful handling of negative signs is essential when substituting the remaining integral.
Always verify complicated indefinite integrals by differentiation.
Do not forget the arbitrary constant \(C\).

📘 Concept & Theory

Theory and Concept

›

The given integrand is the product of an algebraic function \(x\) and an inverse trigonometric function \(\tan^{-1}x\). Integration by parts is suitable because differentiating the inverse tangent function produces a simpler rational expression.

The integration by parts formula is:

\[ \boxed{\int u\,dv=uv-\int v\,du} \]

ILATE Rule

ILATE provides a priority order for selecting the function to be differentiated:

I: Inverse trigonometric functions
L: Logarithmic functions
A: Algebraic functions
T: Trigonometric functions
E: Exponential functions

Since \(\tan^{-1}x\) is an inverse trigonometric function and \(x\) is an algebraic function, ILATE suggests choosing \(\tan^{-1}x\) as \(u\).

Important derivative formula:

\[ \frac{d}{dx}(\tan^{-1}x)=\frac{1}{1+x^2} \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Apply the ILATE rule to select \(u=\tan^{-1}x\) and \(dv=x\,dx\).
Differentiate \(u\) and integrate \(dv\).
Apply integration by parts to obtain a rational integral.
Simplify the rational expression using \(x^2=(1+x^2)-1\).
Evaluate the resulting integrals using standard formulas.
Substitute the result and simplify the final expression.
Verify the answer by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 17 steps

Given —
\[I=\int x\tan^{-1}x\,dx\]
Using the ILATE rule, choose:
\[u=\tan^{-1}x,\quad dv=x\,dx\]
Differentiate \(u\).
\[du=\frac{d}{dx}(\tan^{-1}x)\,dx\]
Using the standard derivative formula:
\[\boxed{du=\frac{1}{1+x^2}\,dx}\]
Integrate \(dv\)
\[v=\int x\,dx\]
Applying the power rule:
\[\boxed{v=\frac{x^2}{2}}\]
Therefore, we have:
\[u=\tan^{-1}x,\quad du=\frac{1}{1+x^2}\,dx,\quad v=\frac{x^2}{2}\]
Apply integration by parts
Recall:
\[\int u\,dv=uv-\int v\,du\]
Substituting the values:
\[\begin{aligned}I&=uv-\int v\,du\\ &=\frac{x^2}{2}\tan^{-1}x-\int\frac{x^2}{2}\cdot\frac{1}{1+x^2}\,dx\\ &=\frac{x^2}{2}\tan^{-1}x-\frac{1}{2}\int\frac{x^2}{1+x^2}\,dx\end{aligned}\]
Let the remaining integral be:
\[J=\int\frac{x^2}{1+x^2}\,dx\]
Simplify the rational expression
We know that:
\[x^2=(1+x^2)-1\]
Therefore:
\[\begin{aligned}J&=\int\frac{(1+x^2)-1}{1+x^2}\,dx\\ &=\int\left[\frac{1+x^2}{1+x^2}-\frac{1}{1+x^2}\right]dx\\ &=\int\left[1-\frac{1}{1+x^2}\right]dx\\ &=\int 1\,dx-\int\frac{1}{1+x^2}\,dx\end{aligned}\]
Evaluate the integrals
Using the standard formulas:
\[\int 1\,dx=x\]
and
\[\int\frac{1}{1+x^2}\,dx=\tan^{-1}x\]
We obtain:
\[ \begin{aligned} J&=x-\tan^{-1}x \end{aligned} \]
Hence:
\[\boxed{\int\frac{x^2}{1+x^2}\,dx=x-\tan^{-1}x+C}\]
Substitute the result into the original integral
Recall:
\[I=\frac{x^2}{2}\tan^{-1}x-\frac{1}{2}J\]
Substituting the value of \(J\):
\[\begin{aligned}I&=\frac{x^2}{2}\tan^{-1}x-\frac{1}{2}(x-\tan^{-1}x)\\ &=\frac{x^2}{2}\tan^{-1}x-\frac{x}{2}+\frac{1}{2}\tan^{-1}x\end{aligned}\]
Taking \(\frac{1}{2}\tan^{-1}x\) common from the first and third terms:
\[\begin{aligned}I&=\frac{1}{2}\tan^{-1}x(x^2+1)-\frac{x}{2}+C\end{aligned}\]
Therefore, the required result is:
\[\boxed{\displaystyle\int x\tan^{-1}x\,dx=\frac{1}{2}\left[(x^2+1)\tan^{-1}x-x\right]+C}\]

💡 Answer

Final Answer

›

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

🎯 Exam Significance

Exam Significance

›

Application of ILATE: The question tests the correct selection of the inverse trigonometric function as \(u\).
Integration by parts: It assesses the ability to apply the formula systematically to a product of two functions.
Standard integral: Students must recall \(\int\frac{1}{1+x^2}\,dx=\tan^{-1}x+C\).
Algebraic simplification: The identity \(x^2=(1+x^2)-1\) converts the rational expression into two elementary integrals.
Presentation of steps: Showing the intermediate simplification and correct handling of signs makes the solution easier to follow and assess.

Significance for Competitive Entrance Examinations

JEE Main and Advanced: This problem reinforces the use of integration by parts with inverse trigonometric functions.
Algebraic decomposition: Recognising how to split a rational expression into simpler terms is an important integration skill.
Standard integral recognition: The result relies on familiarity with the inverse tangent integral.
Efficient problem-solving: A suitable choice of \(u\) and \(dv\) reduces the complexity of the original integral.
Answer verification: Differentiation provides a reliable way to check the correctness of an indefinite integral.

🚫 Common Mistakes

Common Mistakes

›

Incorrect choice of \(u\): Choose \(\tan^{-1}x\) as \(u\) according to ILATE.
Incorrect derivative: Do not confuse the derivative of \(\tan^{-1}x\) with that of \(\sin^{-1}x\). The correct derivative is \(\frac{1}{1+x^2}\).
Missing factor: When integrating \(x\), remember that \(v=\frac{x^2}{2}\), not \(x^2\).
Incorrect algebraic decomposition: Use \(x^2=(1+x^2)-1\) to simplify the numerator.
Sign error: The negative sign in \(-\frac{1}{2}\int\frac{x^2}{1+x^2}\,dx\) must be retained throughout the calculation.
Incorrect simplification: The integral \(\int\frac{x^2}{1+x^2}\,dx\) equals \(x-\tan^{-1}x+C\), not \(x+\tan^{-1}x+C\).
Missing constant: Include \(C\) in the final answer.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 9 points

Use integration by parts when the integrand is a product of an inverse trigonometric function and an algebraic function.
According to ILATE, choose the inverse trigonometric function as \(u\).
For this problem, take \(u=\tan^{-1}x\) and \(dv=x\,dx\).
Remember that \(\frac{d}{dx}(\tan^{-1}x)=\frac{1}{1+x^2}\).
The identity \(x^2=(1+x^2)-1\) simplifies the remaining rational integral.
The standard result \(\int\frac{1}{1+x^2}\,dx=\tan^{-1}x+C\) is essential.
Pay attention to the negative sign before the remaining integral when applying integration by parts.
Verify the final result by differentiation whenever possible.
Always include the arbitrary constant of integration \(C\).

📘 Concept & Theory

Theory and Concept

›

The given integrand is the product of an algebraic function \(x\) and an inverse trigonometric function \(\cos^{-1}x\). Integration by parts is appropriate because differentiating the inverse cosine function produces a simpler expression.

The integration by parts formula is:

\[ \boxed{\int u\,dv=uv-\int v\,du} \]

ILATE Rule

ILATE provides a priority order for selecting the function to be differentiated:

I: Inverse trigonometric functions
L: Logarithmic functions
A: Algebraic functions
T: Trigonometric functions
E: Exponential functions

Since \(\cos^{-1}x\) is an inverse trigonometric function and \(x\) is an algebraic function, ILATE suggests choosing \(\cos^{-1}x\) as \(u\).

Important derivative formula:

\[ \frac{d}{dx}(\cos^{-1}x) =-\frac{1}{\sqrt{1-x^2}} \]

The negative sign in this derivative is particularly important. When applying integration by parts, it changes the sign of the remaining integral from negative to positive.

🗺️ Solution Roadmap

Step-by-step Plan

›

Apply the ILATE rule and select \(u=\cos^{-1}x\) and \(dv=x\,dx\).
Differentiate \(u\) and integrate \(dv\).
Apply integration by parts, keeping track of the negative derivative of \(\cos^{-1}x\).
Evaluate the remaining integral \(\int\frac{x^2}{\sqrt{1-x^2}}\,dx\).
Use the identity \(x^2=1-(1-x^2)\) to simplify the integrand.
Apply the standard integral formula for \(\sqrt{1-x^2}\).
Substitute the result into the original expression and simplify.
Verify the final answer by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 23 steps

Given —
\[I=\int x\cos^{-1}x\,dx\]
Using the ILATE rule, choose:
\[u=\cos^{-1}x,\quad dv=x\,dx\]
Differentiate \(u\)
\[du=\frac{d}{dx}(\cos^{-1}x)\,dx\]
Using the standard derivative formula:
\[\boxed{du=-\frac{1}{\sqrt{1-x^2}}\,dx}\]
Integrate \(dv\)
\[v=\int x\,dx\]
Applying the power rule:
\[\boxed{v=\frac{x^2}{2}}\]
Therefore:
\[u=\cos^{-1}x,\quad du=-\frac{1}{\sqrt{1-x^2}}\,dx,\quad v=\frac{x^2}{2}\]
Apply integration by parts
Recall:
\[\int u\,dv=uv-\int v\,du\]
Substituting the values:
\[\begin{aligned}I&=uv-\int v\,du\\ &=\frac{x^2}{2}\cos^{-1}x-\int\frac{x^2}{2}\left(-\frac{1}{\sqrt{1-x^2}}\right)dx\\ &=\frac{x^2}{2}\cos^{-1}x+\frac{1}{2}\int\frac{x^2}{\sqrt{1-x^2}}\,dx\end{aligned}\]
Let the remaining integral be:
\[J=\int\frac{x^2}{\sqrt{1-x^2}}\,dx\]
Simplify the remaining integral
We know that:
\[x^2=1-(1-x^2)\]
Substituting this identity into \(J\):
\[\begin{aligned}J&=\int\frac{1-(1-x^2)}{\sqrt{1-x^2}}\,dx\\ &=\int\left[\frac{1}{\sqrt{1-x^2}}-\frac{1-x^2}{\sqrt{1-x^2}}\right]dx\\ &=\int\frac{1}{\sqrt{1-x^2}}\,dx-\int\sqrt{1-x^2}\,dx\end{aligned}\]
Therefore:
\[\boxed{J=\sin^{-1}x-\int\sqrt{1-x^2}\,dx}\]
Evaluate the standard integral
We use the standard formula:
\[\boxed{\int\sqrt{a^2-x^2}\,dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)+C}\]
For the given integral, \(a=1\). Therefore:
\[\begin{aligned}\int\sqrt{1-x^2}\,dx&=\frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}\sin^{-1}x\\ &=\frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}\left(\frac{\pi}{2}-\cos^{-1}x\right)\end{aligned}\]
Here, we have used the identity:
\[\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\]
Hence:
\[\boxed{\int\sqrt{1-x^2}\,dx=\frac{x}{2}\sqrt{1-x^2}-\frac{1}{2}\cos^{-1}x+\frac{\pi}{4}+C}\]
Substitute the standard integral into \(J\)
Recall:
\[J=\sin^{-1}x-\int\sqrt{1-x^2}\,dx\]
Substituting the evaluated integral:
\[\begin{aligned}J&=\sin^{-1}x-\left[\frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}\sin^{-1}x\right]\\ &=\frac{1}{2}\sin^{-1}x-\frac{x}{2}\sqrt{1-x^2}\\ &=\frac{1}{2}\left(\frac{\pi}{2}-\cos^{-1}x\right)-\frac{x}{2}\sqrt{1-x^2}\\ &=\frac{\pi}{4}-\frac{1}{2}\cos^{-1}x-\frac{x}{2}\sqrt{1-x^2}\end{aligned}\]
Therefore:
\[\boxed{J=\frac{\pi}{4}-\frac{1}{2}\cos^{-1}x-\frac{x}{2}\sqrt{1-x^2}}\]
Substitute \(J\) into the original integral
\[I=\frac{x^2}{2}\cos^{-1}x+\frac{1}{2}J\]
Substituting the value of \(J\):
\[\begin{aligned}I={}&\frac{x^2}{2}\cos^{-1}x\\ &+\frac{1}{2}\left[\frac{\pi}{4}-\frac{1}{2}\cos^{-1}x-\frac{x}{2}\sqrt{1-x^2}\right]\\ ={}&\frac{x^2}{2}\cos^{-1}x+\frac{\pi}{8}-\frac{1}{4}\cos^{-1}x-\frac{x}{4}\sqrt{1-x^2}\end{aligned}\]
Combining the inverse cosine terms:
\[\begin{aligned}I={}&\left(\frac{x^2}{2}-\frac{1}{4}\right)\cos^{-1}x-\frac{x}{4}\sqrt{1-x^2}+\frac{\pi}{8}+C\\ ={}&\frac{1}{4}(2x^2-1)\cos^{-1}x-\frac{x}{4}\sqrt{1-x^2}+\frac{\pi}{8}+C\end{aligned}\]
Since \(\frac{\pi}{8}\) is a constant, it can be absorbed into the arbitrary constant of integration. Thus, the required result is:
\[\boxed{\displaystyle\int x\cos^{-1}x\,dx=\frac{1}{4}\left[(2x^2-1)\cos^{-1}x-x\sqrt{1-x^2}\right]+C}\]

💡 Answer

Final Answer

›

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

🎯 Exam Significance

Exam Significance

›

Application of ILATE: This problem tests the selection of an inverse trigonometric function as \(u\) when applying integration by parts.
Inverse trigonometric derivatives: Students must remember the negative sign in the derivative of \(\cos^{-1}x\).
Standard integral formula: The problem requires the integral of \(\sqrt{1-x^2}\).
Algebraic manipulation: The identity \(x^2=1-(1-x^2)\) helps simplify the remaining integral.
Trigonometric identity: The relationship \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\) allows the result to be expressed entirely in terms of \(\cos^{-1}x\).
Step-wise marking: Clearly showing the sign changes and intermediate calculations makes the solution easier to follow and assess.

Significance for Competitive Entrance Examinations

JEE Main and Advanced: The problem develops the ability to combine integration by parts with inverse trigonometric functions.
Sign management: The negative derivative of \(\cos^{-1}x\) makes this a useful exercise in handling signs accurately.
Standard integral recognition: Recognising the integral of \(\sqrt{1-x^2}\) is useful in more complex integration problems.
Identity-based simplification: Converting between \(\sin^{-1}x\) and \(\cos^{-1}x\) provides flexibility in presenting the final expression.
Verification skills: Differentiation provides an effective way to check the correctness of an antiderivative.

🚫 Common Mistakes

Common Mistakes

›

Incorrect derivative: Do not omit the negative sign in \(\frac{d}{dx}(\cos^{-1}x)\).
Incorrect integration by parts sign: Carefully apply the minus sign in \(uv-\int v\,du\). The negative derivative of \(\cos^{-1}x\) makes the remaining integral positive.
Incorrect algebraic decomposition: Use \(x^2=1-(1-x^2)\) to obtain the correct difference of two integrals.
Sign error in the standard integral: When rewriting \(\sin^{-1}x\) as \(\frac{\pi}{2}-\cos^{-1}x\), retain the negative sign before \(\cos^{-1}x\).
Incorrect final simplification: The square-root term in the final answer carries a negative sign.
Confusing inverse functions: Remember that \(\sin^{-1}x\) and \(\cos^{-1}x\) are related by their sum being \(\frac{\pi}{2}\), not by equality.
Missing constant: Include the arbitrary constant \(C\) in the final answer.
Domain restriction: The real-valued derivative formula applies for \(-1

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 9 points

Use integration by parts for products involving inverse trigonometric and algebraic functions.
According to ILATE, choose \(\cos^{-1}x\) as \(u\).
Remember that \(\frac{d}{dx}(\cos^{-1}x)=-\frac{1}{\sqrt{1-x^2}}\).
The negative derivative changes the sign of the remaining integral when applying integration by parts.
The identity \(x^2=1-(1-x^2)\) simplifies the remaining integral.
The standard integral of \(\sqrt{1-x^2}\) is essential for completing the calculation.
Use \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\) to express the result in terms of inverse cosine.
A constant such as \(\frac{\pi}{8}\) can be absorbed into the arbitrary integration constant.
Verify the final answer by differentiation.

📘 Concept & Theory

Theory and Concept

›

The given integral contains the square of an inverse trigonometric function. Since it is not directly reducible to a standard integration formula, we use integration by parts.

The integration by parts formula is:

\[ \boxed{\int u\,dv=uv-\int v\,du} \]

The key idea is to differentiate \((\sin^{-1}x)^2\) and integrate \(1\). Differentiating the inverse sine function introduces the factor \(\frac{1}{\sqrt{1-x^2}}\), which can be simplified by using the substitution \(x=\sin\theta\).

Important derivative formula:

\[ \frac{d}{dx}(\sin^{-1}x) =\frac{1}{\sqrt{1-x^2}} \]

🗺️ Solution Roadmap

Step-by-step Plan

›

Apply integration by parts by choosing \(u=(\sin^{-1}x)^2\) and \(dv=dx\).
Differentiate \(u\) using the chain rule and integrate \(dv\).
Express the resulting integral in terms of \(\sin^{-1}x\).
Use the substitution \(x=\sin\theta\) to simplify the remaining integral.
Apply integration by parts again to evaluate the trigonometric integral.
Substitute back \(\theta=\sin^{-1}x\) and simplify.
Verify the final result by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 27 steps

Given —
\[I=\int(\sin^{-1}x)^2\,dx\]
Choose:
\[u=(\sin^{-1}x)^2,\qquad dv=dx\]
Differentiate \(u\) and integrate \(dv\)
Using the chain rule:
\[\begin{aligned}du&=\frac{d}{dx}(\sin^{-1}x)^2\,dx\\ &=2\sin^{-1}x\cdot\frac{d}{dx}(\sin^{-1}x)\,dx\\ &=\frac{2\sin^{-1}x}{\sqrt{1-x^2}}\,dx\end{aligned}\]
Also:
\[v=\int dx=x\]
Therefore:
\[u=(\sin^{-1}x)^2,\quad du=\frac{2\sin^{-1}x}{\sqrt{1-x^2}}\,dx,\quad v=x\]
Substitute into the integration by parts formula.
Using:
\[\int u\,dv=uv-\int v\,du\]
We obtain:
\[\begin{aligned}I&=x(\sin^{-1}x)^2-\int x\cdot\frac{2\sin^{-1}x}{\sqrt{1-x^2}}\,dx\\ &=x(\sin^{-1}x)^2-2\int\frac{x\sin^{-1}x}{\sqrt{1-x^2}}\,dx\end{aligned}\]
Let:
\[J=\int\frac{x\sin^{-1}x}{\sqrt{1-x^2}}\,dx\]
Thus:
\[\boxed{I=x(\sin^{-1}x)^2-2J}\]
Evaluate \(J\) using substitution.
Put:
\[x=\sin\theta\]
Differentiating both sides:
\[dx=\cos\theta\,d\theta\]
Also:
\[\sqrt{1-x^2}=\sqrt{1-\sin^2\theta}=\cos\theta\]
For the principal inverse sine range, \(-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}\), we have \(\cos\theta\geq0\), so the positive square root is \(\cos\theta\).
Further:
\[\sin^{-1}x=\theta\]
Substituting these values into \(J\):
\[\begin{aligned}J&=\int\frac{\sin\theta\cdot\theta}{\cos\theta}\cos\theta\,d\theta\\ &=\int\theta\sin\theta\,d\theta\end{aligned}\]
Therefore:
\[\boxed{J=\int\theta\sin\theta\,d\theta}\]
Evaluate the trigonometric integral using integration by parts
We have:
\[J=\int\theta\sin\theta\,d\theta\]
Choose:
\[u=\theta,\qquad dv=\sin\theta\,d\theta\]
Then:
\[du=d\theta,\qquad v=-\cos\theta\]
Applying integration by parts:
\[\begin{aligned}J&=uv-\int v\,du\\ &=-\theta\cos\theta+\int\cos\theta\,d\theta\\ &=-\theta\cos\theta+\sin\theta+C\end{aligned}\]
Hence:
\[\boxed{J=\sin\theta-\theta\cos\theta+C}\]
Substitute back in terms of \(x\)
Recall:
\[x=\sin\theta,\quad \theta=\sin^{-1}x\]
Also:
\[\cos\theta=\sqrt{1-x^2}\]
Therefore:
\[\begin{aligned}J&=\sin\theta-\theta\cos\theta\\ &=x-\sin^{-1}x\sqrt{1-x^2}\end{aligned}\]
Thus:
\[\boxed{J=x-\sin^{-1}x\sqrt{1-x^2}}\]
Substitute \(J\) into the original integral
We obtained:
\[I=x(\sin^{-1}x)^2-2J\]
Substituting the value of \(J\):
\[\begin{aligned}I&=x(\sin^{-1}x)^2-2\left[x-\sin^{-1}x\sqrt{1-x^2}\right]\\ &=x(\sin^{-1}x)^2-2x+2\sin^{-1}x\sqrt{1-x^2}\end{aligned}\]
Therefore, the required result is:
\[\boxed{\displaystyle\int(\sin^{-1}x)^2\,dx=x(\sin^{-1}x)^2+2\sin^{-1}x\sqrt{1-x^2}-2x+C}\]

💡 Answer

Final Answer

›

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

🎯 Exam Significance

Exam Significance

›

Integration by parts: This problem tests the application of integration by parts to the square of an inverse trigonometric function.
Chain rule: Students must correctly differentiate \((\sin^{-1}x)^2\).
Trigonometric substitution: The substitution \(x=\sin\theta\) simplifies the radical expression \(\sqrt{1-x^2}\).
Repeated integration by parts: The problem requires integration by parts twice, first in \(x\) and then in \(\theta\).
Step-wise presentation: Showing the substitution, simplification and back-substitution clearly demonstrates the complete method.

Significance for Competitive Entrance Examinations

JEE Main and Advanced: This problem combines integration by parts, inverse trigonometric derivatives and trigonometric substitution.
Strategic substitution: Recognising that \(x=\sin\theta\) eliminates the radical and converts the inverse sine into a simple variable.
Multi-stage integration: It develops the ability to select an appropriate technique at each stage of a problem.
Conceptual understanding: The problem demonstrates how inverse trigonometric expressions can be transformed into ordinary trigonometric integrals.
Verification: Differentiation confirms the result and helps identify errors in signs or coefficients.

🚫 Common Mistakes

Common Mistakes

›

Incorrect differentiation: Remember that
\[ \frac{d}{dx}(\sin^{-1}x)^2 =\frac{2\sin^{-1}x}{\sqrt{1-x^2}}. \]
Missing factor of 2: The chain rule produces the factor \(2\sin^{-1}x\).
Incorrect substitution: When \(x=\sin\theta\), use \(dx=\cos\theta\,d\theta\) and \(\sqrt{1-x^2}=\cos\theta\) on the principal inverse sine interval.
Sign error: Remember that \(\int\theta\sin\theta\,d\theta=-\theta\cos\theta+\sin\theta+C\).
Incomplete back-substitution: Replace both \(\theta\) and \(\cos\theta\) with their expressions in terms of \(x\).
Missing factor: Do not forget the factor \(-2\) multiplying the remaining integral after the first integration by parts.
Missing constant: Include \(C\) in the final answer.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 9 points

For the integral of \((\sin^{-1}x)^2\), integration by parts is an effective first step.
Choose \(u=(\sin^{-1}x)^2\) and \(dv=dx\).
Apply the chain rule carefully when differentiating the square of an inverse trigonometric function.
The substitution \(x=\sin\theta\) simplifies \(\sqrt{1-x^2}\) to \(\cos\theta\) within the principal inverse sine range.
After substitution, the remaining integral becomes \(\int\theta\sin\theta\,d\theta\).
A second application of integration by parts evaluates the trigonometric integral.
Always substitute back into the original variable \(x\).
Verify the final result by differentiation.
Include the arbitrary constant of integration \(C\).

📘 Concept & Theory

Theory and Concept

›

The given integrand contains an inverse trigonometric function multiplied by an algebraic expression involving a square root. We can solve this integral efficiently by combining the substitution method with integration by parts.

The integration by parts formula is:

\[ \boxed{\int u\,dv=uv-\int v\,du} \]

The key observation is that the integral

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

can be evaluated easily using the substitution \(t=1-x^2\). Its result is particularly convenient because it cancels the square-root expression in the derivative of \(\cos^{-1}x\).

Important derivative formula:

\[ \frac{d}{dx}(\cos^{-1}x) =-\frac{1}{\sqrt{1-x^2}} \]

🗺️ Solution Roadmap

Step-by-step Plan

›

First evaluate \(\int\frac{x}{\sqrt{1-x^2}}\,dx\) using substitution.
Choose \(u=\cos^{-1}x\) and \(dv=\frac{x}{\sqrt{1-x^2}}\,dx\).
Calculate \(du\) and \(v\).
Apply integration by parts.
Simplify the resulting integral and obtain the final expression.
Verify the result by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 16 steps

Evaluate the simpler integral using substitution
Consider:
\[J=\int\frac{x}{\sqrt{1-x^2}}\,dx\]
Put:
\[t=1-x^2\]
Differentiating both sides:
\[dt=-2x\,dx\]
Therefore:
\[x\,dx=-\frac{1}{2}dt\]
Substituting into the integral:
\[\begin{aligned}J&=\int\frac{x}{\sqrt{1-x^2}}\,dx\\ &=-\frac{1}{2}\int\frac{1}{\sqrt{t}}\,dt\\ &=-\frac{1}{2}\int t^{-1/2}\,dt\end{aligned}\]
Using the power rule of integration:
\[\begin{aligned}J&=-\frac{1}{2}\left(\frac{t^{1/2}}{1/2}\right)+C\\ &=-\sqrt{t}+C\end{aligned}\]
Substituting \(t=1-x^2\):
\[\boxed{J=-\sqrt{1-x^2}+C}\]
Important: The negative sign is essential because the derivative of \(1-x^2\) is \(-2x\).
Identify the functions for integration by parts
The given integral is:
\[I=\int\frac{x\cos^{-1}x}{\sqrt{1-x^2}}\,dx\]
Using the integration by parts formula, choose:
\[u=\cos^{-1}x,\quad dv=\frac{x}{\sqrt{1-x^2}}\,dx\]
Calculate \(du\) and \(v\)
Differentiating \(u\):
\[du=-\frac{1}{\sqrt{1-x^2}}\,dx\]
Integrating \(dv\),
\[v=\int\frac{x}{\sqrt{1-x^2}}\,dx=-\sqrt{1-x^2}\]
Thus:
\[\boxed{u=\cos^{-1}x,\quad du=-\frac{1}{\sqrt{1-x^2}}\,dx,\quad v=-\sqrt{1-x^2}}\]
Apply integration by parts
Recall:
\[\int u\,dv=uv-\int v\,du\]
Substituting the values:
\[\begin{aligned}I&=uv-\int v\,du\\ &=-\cos^{-1}x\sqrt{1-x^2}\\ &\quad-\int\left(-\sqrt{1-x^2}\right)\left(-\frac{1}{\sqrt{1-x^2}}\right)dx\\ &=-\cos^{-1}x\sqrt{1-x^2}-\int 1\,dx\end{aligned}\]
Simplifying:
\[I=-\cos^{-1}x\sqrt{1-x^2}-x+C\]
Therefore, the required result is:
\[\boxed{\displaystyle\int\frac{x\cos^{-1}x}{\sqrt{1-x^2}}\,dx=-x-\cos^{-1}x\sqrt{1-x^2}+C}\]

💡 Answer

Final Answer

›

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

🎯 Exam Significance

Exam Significance

›

Combination of methods: This problem tests the application of substitution followed by integration by parts.
Inverse trigonometric derivatives: Students must remember the negative sign in the derivative of \(\cos^{-1}x\).
Substitution technique: The substitution \(t=1-x^2\) simplifies an integral containing a square-root expression.
Integration by parts: The question demonstrates how a carefully selected \(u\) and \(dv\) simplify a complicated product.
Sign accuracy: Correctly handling the two negative signs in \(v\) and \(du\) is essential.
Step-wise marking: Showing the preliminary integral separately makes the complete method clear.

Significance for Competitive Entrance Examinations

JEE Main and Advanced: This problem develops proficiency in combining substitution and integration by parts.
Strategic selection: Recognising that the derivative of \(\cos^{-1}x\) contains the same square-root expression as the integrand is an important problem-solving insight.
Algebraic simplification: The cancellation of \(\sqrt{1-x^2}\) produces a simple integral, reducing the calculation significantly.
Sign management: The problem reinforces careful handling of negative signs in inverse trigonometric differentiation and integration by parts.
Verification skills: Differentiation provides a reliable method to confirm the final antiderivative.

🚫 Common Mistakes

Common Mistakes

›

Incorrect substitution sign: If \(t=1-x^2\), then \(dt=-2x\,dx\), not \(2x\,dx\).
Incorrect preliminary integral: Remember that \(\int\frac{x}{\sqrt{1-x^2}}\,dx=-\sqrt{1-x^2}+C\).
Incorrect derivative: The derivative of \(\cos^{-1}x\) is negative.
Sign error in integration by parts: Apply \(uv-\int v\,du\) exactly as written. Do not lose the negative sign before the remaining integral.
Incorrect multiplication: The product of \(v=-\sqrt{1-x^2}\) and \(du=-\frac{1}{\sqrt{1-x^2}}dx\) is \(+1\,dx\).
Incorrect final sign: Both the \(x\) term and the \(\cos^{-1}x\sqrt{1-x^2}\) term carry negative signs.
Missing constant: Include \(C\) in the final indefinite integral.
Domain restriction: The real-valued inverse cosine and square-root expressions are defined for \(-1\leq x\leq1\); the differentiation verification applies on \(-1

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 8 points

Evaluate the simpler integral before applying integration by parts when it makes the calculation easier.
For \(\int\frac{x}{\sqrt{1-x^2}}\,dx\), use the substitution \(t=1-x^2\).
Remember that the correct result is \(-\sqrt{1-x^2}+C\).
For the original integral, choose \(u=\cos^{-1}x\) and \(dv=\frac{x}{\sqrt{1-x^2}}\,dx\).
The derivative of \(\cos^{-1}x\) contains a negative sign.
The product of the two negative factors in \(v\,du\) is positive, but the integration by parts formula has an additional minus sign.
Always verify the final result by differentiation.
Include the arbitrary constant of integration \(C\).

📘 Concept & Theory

Theory and Concept

›

The given integrand is the product of an algebraic function \(x\) and a trigonometric function \(\sec^2x\). According to the ILATE rule, the algebraic function is selected for differentiation, while the trigonometric function is integrated.

The integration by parts formula is:

\[ \boxed{\int u\,dv=uv-\int v\,du} \]

We use the standard integral:

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

The remaining integral involves \(\tan x\), which can be evaluated using the substitution \(t=\cos x\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Apply the ILATE rule and choose \(u=x\) and \(dv=\sec^2x\,dx\).
Differentiate \(u\) and integrate \(dv\).
Apply integration by parts to obtain an integral involving \(\tan x\).
Rewrite \(\tan x\) as \(\frac{\sin x}{\cos x}\).
Use the substitution \(t=\cos x\) to evaluate the remaining integral.
Substitute back and simplify the final expression.
Verify the result by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 19 steps

Given —
\[I=\int x\sec^2x\,dx\]
Identify the functions using ILATE.
According to the ILATE rule, choose:
\[u=x,\quad dv=\sec^2x\,dx\]
Calculate \(du\) and \(v\)
Differentiating \(u=x\):
\[du=dx\]
Integrating \(dv\):
\[ \begin{aligned} v&=\int\sec^2x\,dx\\ &=\tan x \end{aligned} \]
Therefore:
\[\boxed{u=x,\quad du=dx,\quad v=\tan x}\]
Apply integration by parts
Using the formula:
\[\int u\,dv=uv-\int v\,du\]
Substituting the values:
\[\begin{aligned}I&=uv-\int v\,du\\ &=x\tan x-\int\tan x\,dx\end{aligned}\]
Thus, the original integral becomes:
\[\boxed{I=x\tan x-\int\tan x\,dx}\]
Rewrite the tangent function
Using the trigonometric identity: We obtain:
\[\begin{aligned}I&=x\tan x-\int\frac{\sin x}{\cos x}\,dx\end{aligned}\]
Evaluate the remaining integral using substitution.
Consider:
\[J=\int\frac{\sin x}{\cos x}\,dx\]
Put:
\[t=\cos x\]
Differentiating both sides:
\[dt=-\sin x\,dx\]
Therefore:
\[\sin x\,dx=-dt\]
Substituting into the integral:
\[ \begin{aligned} J&=\int\frac{\sin x}{\cos x}\,dx\\ &=-\int\frac{1}{t}\,dt\\ &=-\ln|t|+C \end{aligned} \]
Substituting \(t=\cos x\):
\[\boxed{J=-\ln|\cos x|+C}\]
Substitute the result into the original integral
Recall:
\[I=x\tan x-\int\tan x\,dx\]
Since:
\[\int\tan x\,dx=-\ln|\cos x|+C\]
We obtain:
\[ \begin{aligned} I&=x\tan x-\left[-\ln|\cos x|\right]+C\\ &=x\tan x+\ln|\cos x|+C \end{aligned} \]
Therefore, the required result is:
\[\boxed{\displaystyle\int x\sec^2x\,dx=x\tan x+\ln|\cos x|+C}\]
Equivalent form:
Since \(\ln|\cos x|=-\ln|\sec x|\), the answer can also be written as:
\[\boxed{\int x\sec^2x\,dx=x\tan x-\ln|\sec x|+C}\]

💡 Answer

Final Answer

›

\[\boxed{\displaystyle\int x\sec^2x\,dx=x\tan x+\ln|\cos x|+C}\]

🎯 Exam Significance

Exam Significance

›

Application of ILATE: The problem tests the selection of the algebraic function as \(u\) and the trigonometric function as \(dv\).
Integration by parts: Students must correctly apply the formula to a product of algebraic and trigonometric functions.
Standard trigonometric integral: The result \(\int\sec^2x\,dx=\tan x+C\) is essential.
Substitution method: The substitution \(t=\cos x\) simplifies the integral of \(\tan x\).
Logarithmic integration: The problem reinforces the formula \(\int\frac{1}{t}\,dt=\ln|t|+C\).
Sign accuracy: Correctly handling the negative sign in \(dt=-\sin x\,dx\) is important for obtaining the correct final answer.

Significance for Competitive Entrance Examinations

JEE Main and Advanced: This problem develops proficiency in combining integration by parts with substitution.
Technique selection: Recognising the suitable choice of \(u\) and \(dv\) helps reduce the complexity of integration problems.
Trigonometric manipulation: Converting tangent into a sine-cosine ratio makes the substitution method straightforward.
Logarithmic forms: Students learn to recognise equivalent logarithmic expressions involving \(\cos x\) and \(\sec x\).
Verification skills: Differentiation confirms the result and helps detect errors in signs and coefficients.

🚫 Common Mistakes

Common Mistakes

›

Incorrect choice of \(u\): Choose \(u=x\) and \(dv=\sec^2x\,dx\) according to ILATE.
Incorrect integration: Remember that the integral of \(\sec^2x\) is \(\tan x\), not \(\sec x\).
Substitution sign error: If \(t=\cos x\), then \(dt=-\sin x\,dx\).
Incorrect tangent integral: The correct result is \(\int\tan x\,dx=-\ln|\cos x|+C\).
Sign error: The negative sign before \(\int\tan x\,dx\) in the integration by parts result changes the negative logarithm into a positive logarithmic term.
Missing absolute-value signs: Write \(\ln|\cos x|\), not simply \(\ln(\cos x)\), for the real-valued antiderivative on intervals where \(\cos x\neq0\).
Missing constant: Include the arbitrary constant of integration \(C\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 9 points

For \(\int x\sec^2x\,dx\), integration by parts is the most convenient method.
According to ILATE, choose \(u=x\) and \(dv=\sec^2x\,dx\).
Remember that \(\int\sec^2x\,dx=\tan x+C\).
The remaining integral is \(\int\tan x\,dx\).
Use \(t=\cos x\) to evaluate the integral of \(\tan x\).
The standard result is \(\int\tan x\,dx=-\ln|\cos x|+C\).
Therefore, subtracting this integral produces a positive logarithmic term in the final answer.
Always use absolute-value signs in the logarithmic expression.
Verify the final result by differentiation.

📘 Concept & Theory

Theory and Concept

›

The given integral contains the inverse trigonometric function \(\tan^{-1}x\). Since it is not directly integrable using a standard integration formula, we apply integration by parts.

According to the ILATE rule, inverse trigonometric functions are given priority over algebraic functions when selecting the first function for differentiation.

The integration by parts formula is:

\[ \boxed{\int u\,dv=uv-\int v\,du} \]

We also use the standard derivative:

\[ \frac{d}{dx}(\tan^{-1}x)=\frac{1}{1+x^2} \]

After applying integration by parts, the remaining integral can be evaluated using the substitution \(t=1+x^2\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Rewrite the integral as \(\int 1\cdot\tan^{-1}x\,dx\).
Apply the ILATE rule and select \(u=\tan^{-1}x\) and \(dv=dx\).
Differentiate \(u\) and integrate \(dv\).
Apply integration by parts to obtain an algebraic rational integral.
Use the substitution \(t=1+x^2\) to evaluate the remaining integral.
Substitute back and simplify the result.
Verify the answer by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 16 steps

Given — Rewrite the integral
\[ I=\int\tan^{-1}x\,dx =\int 1\cdot\tan^{-1}x\,dx \]
Select the functions using ILATE
According to the ILATE rule, choose:
\[u=\tan^{-1}x,\quad dv=dx\]
Calculate \(du\) and \(v\)
Differentiating \(u\):
\[du=\frac{1}{1+x^2}\,dx\]
Integrating \(dv\):
\[v=\int dx=x\]
Therefore:
\[\boxed{u=\tan^{-1}x,\quad du=\frac{dx}{1+x^2},\quad v=x}\]
Apply integration by parts
Using the formula:
\[\int u\,dv=uv-\int v\,du\]
Substituting the values:
\[\begin{aligned}I &=x\tan^{-1}x-\int x\left(\frac{1}{1+x^2}\right)dx\\ &=x\tan^{-1}x-\int\frac{x}{1+x^2}\,dx \end{aligned} \]
Thus, we obtain:
\[\boxed{I=x\tan^{-1}x-\int\frac{x}{1+x^2}\,dx}\]
Evaluate the remaining integral using substitution
Consider:
\[J=\int\frac{x}{1+x^2}\,dx\]
Put:
\[t=1+x^2\]
Differentiating both sides with respect to \(x\):
\[\frac{dt}{dx}=2x\]
Therefore:
\[dt=2x\,dx\]
Or:
\[x\,dx=\frac{dt}{2}\]
Substituting \(t=1+x^2\) and \(x\,dx=\frac{dt}{2}\) into the integral:
\[ \begin{aligned} J &=\int\frac{x}{1+x^2}\,dx\\ &=\frac12\int\frac{1}{t}\,dt\\ &=\frac12\ln|t|+C\\ &=\frac12\ln|1+x^2|+C \end{aligned} \]
Since \(1+x^2\) is strictly positive for every real value of \(x\), the absolute-value signs can be omitted:
\[\boxed{J=\frac12\ln(1+x^2)+C}\]
Substitute the result into the original integral.
Recall:
\[I=x\tan^{-1}x-\int\frac{x}{1+x^2}\,dx\]
Substituting the value of the remaining integral:
\[\begin{aligned}I&=x\tan^{-1}x-\frac12\ln(1+x^2)+C\end{aligned}\]
Hence, the required integral is:
\[\boxed{\displaystyle\int\tan^{-1}x\,dx=x\tan^{-1}x-\frac12\ln(1+x^2)+C}\]

💡 Answer

Final Answer

›

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

🎯 Exam Significance

Exam Significance

›

Application of ILATE: The problem tests the correct selection of an inverse trigonometric function for differentiation.
Integration by parts: Students must correctly apply the formula to an inverse trigonometric function multiplied by unity.
Derivative of inverse trigonometric functions: The standard derivative of \(\tan^{-1}x\) is essential for solving this integral.
Substitution method: The substitution \(t=1+x^2\) simplifies the resulting rational integral.
Logarithmic integration: The problem reinforces the standard result \(\int\frac{1}{t}\,dt=\ln|t|+C\).
Stepwise presentation: Showing the choice of \(u\), \(dv\), \(du\), and \(v\) makes the method clear and supports the award of stepwise marks.

Significance for Competitive Entrance Examinations

JEE Main and Advanced: This is a standard application of integration by parts involving inverse trigonometric functions.
Method selection: Recognising when to apply integration by parts is an important problem-solving skill.
Algebraic substitution: The substitution \(t=1+x^2\) demonstrates how to simplify rational expressions efficiently.
Derivative-integral connection: The relationship between the derivative of \(\tan^{-1}x\) and the logarithmic integral is useful in more complex integration problems.
Answer verification: Differentiation provides a reliable method for checking the correctness of an antiderivative.

🚫 Common Mistakes

Common Mistakes

›

Incorrect selection of functions: Choose \(u=\tan^{-1}x\) and \(dv=dx\), rather than differentiating the constant function \(1\).
Incorrect derivative: Remember that the derivative of \(\tan^{-1}x\) is \(\frac{1}{1+x^2}\), not \(\frac{1}{\sqrt{1-x^2}}\).
Incorrect substitution: If \(t=1+x^2\), then \(dt=2x\,dx\), so \(x\,dx=\frac{dt}{2}\).
Missing coefficient: Do not omit the factor \(\frac12\) obtained from the substitution.
Sign error: The integration by parts formula contains a negative sign before the remaining integral. This produces the negative logarithmic term in the final answer.
Unnecessary absolute-value notation: Since \(1+x^2\) is always positive for real \(x\), writing \(\ln(1+x^2)\) is sufficient.
Missing constant: Include the arbitrary constant of integration \(C\).

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 8 points

Rewrite \(\int\tan^{-1}x\,dx\) as \(\int 1\cdot\tan^{-1}x\,dx\) before applying integration by parts.
According to ILATE, select \(u=\tan^{-1}x\) and \(dv=dx\).
Remember that \(\frac{d}{dx}(\tan^{-1}x)=\frac{1}{1+x^2}\).
The integration by parts formula produces the remaining integral \(\int\frac{x}{1+x^2}\,dx\).
Use \(t=1+x^2\), so that \(dt=2x\,dx\).
Since \(1+x^2>0\) for all real \(x\), \(\ln|1+x^2|=\ln(1+x^2)\).
Always retain the negative sign before the logarithmic term when substituting the remaining integral into the integration by parts result.
Differentiate the final expression to verify the answer.

📘 Concept & Theory

Theory and Concept

›

The given integral contains the product of an algebraic function \(x\) and the square of a logarithmic function \((\log x)^2\). Since the logarithmic function becomes simpler upon differentiation, integration by parts is an appropriate method.

According to the ILATE rule, logarithmic functions are selected before algebraic functions for differentiation. Therefore, we take:

\[ u=(\log x)^2,\qquad dv=x\,dx \]

The integration by parts formula is:

\[ \boxed{\int u\,dv=uv-\int v\,du} \]

The first application of integration by parts reduces the power of the logarithmic function from \(2\) to \(1\). We then apply integration by parts again to evaluate the remaining integral.

🗺️ Solution Roadmap

Step-by-step Plan

›

Apply the ILATE rule and choose \(u=(\log x)^2\) and \(dv=x\,dx\).
Calculate \(du\) and \(v\).
Apply integration by parts to reduce the original integral to one involving \(x\log x\).
Evaluate the remaining integral using integration by parts again.
Substitute the second result into the first and simplify.
Verify the final expression by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 15 steps

Given —
\[I=\int x(\log x)^2\,dx\]
Identify the functions using ILATE
According to the ILATE rule, choose:
\[u=(\log x)^2,\quad dv=x\,dx\]
Calculate \(du\) and \(v\).
Differentiating \(u\) using the chain rule:
\[ \begin{aligned} du&=\frac{d}{dx}(\log x)^2\,dx\\ &=2\log x\cdot\frac{1}{x}\,dx\\ &=\frac{2\log x}{x}\,dx \end{aligned} \]
Integrating \(dv\):
\[ \begin{aligned} v&=\int x\,dx\\ &=\frac{x^2}{2} \end{aligned} \]
Therefore:
\[\boxed{u=(\log x)^2,\quad du=\frac{2\log x}{x}\,dx,\quad v=\frac{x^2}{2}}\]
Apply integration by parts for the first time.
Using:
\[\int u\,dv=uv-\int v\,du\]
Substituting the values:
\[ \begin{aligned} I &=\frac{x^2}{2}(\log x)^2 -\int\frac{x^2}{2}\cdot\frac{2\log x}{x}\,dx\\ &=\frac{x^2}{2}(\log x)^2 -\int\frac{x\log x}{1}\,dx\\ &=\frac{x^2}{2}(\log x)^2-\int x\log x\,dx \end{aligned} \]
Thus:
\[\boxed{I=\frac{x^2}{2}(\log x)^2-\int x\log x\,dx}\]
Evaluate the remaining integral using integration by parts again
Let:
\[J=\int x\log x\,dx\]
Applying the ILATE rule, choose:
\[u=\log x,\quad dv=x\,dx\]
Differentiating \(u\):
\[du=\frac{1}{x}\,dx\]
Integrating \(dv\):
\[v=\int x\,dx=\frac{x^2}{2}\]
Applying integration by parts:
\[ \begin{aligned} J &=uv-\int v\,du\\ &=\frac{x^2}{2}\log x -\int\frac{x^2}{2}\cdot\frac{1}{x}\,dx\\ &=\frac{x^2}{2}\log x -\frac12\int x\,dx\\ &=\frac{x^2}{2}\log x -\frac12\left(\frac{x^2}{2}\right)\\ &=\frac{x^2}{2}\log x-\frac{x^2}{4} \end{aligned} \]
Therefore:
\[\boxed{J=\frac{x^2}{2}\log x-\frac{x^2}{4}}\]
Substitute the value of \(J\) into the original integral
Recall:
\[I=\frac{x^2}{2}(\log x)^2-J\]
Substituting the value of \(J\):
\[ \begin{aligned} I &=\frac{x^2}{2}(\log x)^2 -\left(\frac{x^2}{2}\log x-\frac{x^2}{4}\right)\\ &=\frac{x^2}{2}(\log x)^2 -\frac{x^2}{2}\log x+\frac{x^2}{4} \end{aligned} \]
Hence, the required integral is:
\[\boxed{\displaystyle\int x(\log x)^2\,dx=\frac{x^2}{2}(\log x)^2-\frac{x^2}{2}\log x+\frac{x^2}{4}+C}\]

💡 Answer

Final Answer

›

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

🎯 Exam Significance

Exam Significance

›

Repeated integration by parts: This problem tests the application of integration by parts more than once to evaluate an integral involving a squared logarithmic function.
ILATE rule: Students must identify the logarithmic function as the part to be differentiated.
Chain rule: Differentiating \((\log x)^2\) requires the correct application of the chain rule.
Algebraic simplification: Simplifying the product \(\frac{x^2}{2}\cdot\frac{2\log x}{x}\) correctly is essential.
Sign management: The negative sign before the second integral must be retained when substituting the value of \(J\).
Stepwise marking: Clearly showing both applications of integration by parts helps demonstrate the complete method.

Significance for Competitive Entrance Examinations

JEE Main and Advanced: The problem develops proficiency in repeated integration by parts, a frequently used integration technique.
Reduction of logarithmic powers: Each differentiation reduces the power of the logarithmic expression, making the integral progressively simpler.
Product differentiation: The problem reinforces the use of the product rule and chain rule in reverse through integration by parts.
Algebraic accuracy: Correct simplification of coefficients and powers of \(x\) is important in avoiding errors in multistep integration problems.
Verification technique: Differentiating the final expression provides an effective way to check the correctness of a complicated antiderivative.

🚫 Common Mistakes

Common Mistakes

›

Incorrect differentiation: Remember that \(\frac{d}{dx}(\log x)^2=\frac{2\log x}{x}\).
Incorrect simplification: The product \(\frac{x^2}{2}\cdot\frac{2\log x}{x}\) simplifies to \(x\log x\), not \(2x\log x\).
Incorrect second integration: When evaluating \(\int x\log x\,dx\), use \(u=\log x\) and \(dv=x\,dx\).
Sign error during substitution: Subtract the entire expression \(\frac{x^2}{2}\log x-\frac{x^2}{4}\), giving a positive \(\frac{x^2}{4}\) in the final result.
Missing coefficient: Do not omit the factor \(\frac12\) when integrating \(x\).
Domain restriction: Since \(\log x\) is defined for real \(x>0\), state this restriction when discussing the real-valued integral.
Missing constant: Include \(C\) in the final answer.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 8 points

For \(\int x(\log x)^2\,dx\), repeated integration by parts is an effective method.
In the first application, choose \(u=(\log x)^2\) and \(dv=x\,dx\).
Use the chain rule to obtain \(du=\frac{2\log x}{x}\,dx\).
The first application reduces the original integral to \(\frac{x^2}{2}(\log x)^2-\int x\log x\,dx\).
Apply integration by parts again to evaluate \(\int x\log x\,dx\).
Carefully distribute the negative sign when substituting the second integral's value.
For real-valued logarithms, the original integrand is defined for \(x>0\).
Always include the constant of integration \(C\).

📘 Concept & Theory

Theory and Concept

›

The given integrand is the product of an algebraic expression \((x^2+1)\) and a logarithmic function \(\log x\). According to the ILATE rule, logarithmic functions are selected for differentiation, while algebraic functions are integrated.

The integration by parts formula is:

\[ \boxed{\int u\,dv=uv-\int v\,du} \]

In this problem, differentiating \(\log x\) gives \(\frac{1}{x}\), while integrating \(x^2+1\) produces a polynomial expression. This makes integration by parts a convenient method.

🗺️ Solution Roadmap

Step-by-step Plan

›

Apply the ILATE rule to select the logarithmic function for differentiation.
Choose \(u=\log x\) and \(dv=(x^2+1)\,dx\).
Calculate \(du\) and \(v\).
Apply the integration by parts formula.
Simplify and evaluate the remaining polynomial integral.
Combine the terms and verify the result by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 9 steps

Given —
\[I=\int(x^2+1)\log x\,dx\]
According to the ILATE rule, choose:
\[u=\log x,\qquad dv=(x^2+1)\,dx\]
Calculate \(du\) and \(v\)
Differentiating \(u\):
\[du=\frac{1}{x}\,dx\]
Integrating \(dv\):
\[ \begin{aligned} v&=\int(x^2+1)\,dx\\ &=\int x^2\,dx+\int1\,dx\\ &=\frac{x^3}{3}+x \end{aligned} \]
Therefore:
\[\boxed{u=\log x,\quad du=\frac{1}{x}\,dx,\quad v=\frac{x^3}{3}+x}\]
Apply integration by parts
Using the formula:
\[\int u\,dv=uv-\int v\,du\]
Substituting the calculated values:
\[ \begin{aligned} I &=\left(\log x\right)\left(\frac{x^3}{3}+x\right) -\int\left(\frac{x^3}{3}+x\right)\frac{1}{x}\,dx\\ &=\left(\frac{x^3}{3}+x\right)\log x -\int\left(\frac{x^2}{3}+1\right)\,dx \end{aligned} \]
Evaluate the remaining polynomial integral
Integrating each term separately:
\[ \begin{aligned} \int\left(\frac{x^2}{3}+1\right)\,dx &=\frac13\int x^2\,dx+\int1\,dx\\ &=\frac13\cdot\frac{x^3}{3}+x\\ &=\frac{x^3}{9}+x \end{aligned} \]
Substitute and simplify.
Substituting the result into the expression for \(I\):
\[ \begin{aligned} I &=\left(\frac{x^3}{3}+x\right)\log x -\left(\frac{x^3}{9}+x\right)+C\\ &=\frac{x^3}{3}\log x+x\log x -\frac{x^3}{9}-x+C \end{aligned} \]
Hence, the required integral is:
\[\boxed{\displaystyle\int(x^2+1)\log x\,dx=\left(\frac{x^3}{3}+x\right)\log x-\frac{x^3}{9}-x+C}\]

💡 Answer

Final Answer

›

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

🎯 Exam Significance

Exam Significance

›

Application of ILATE: The problem tests the correct selection of the logarithmic function for differentiation.
Integration by parts: Students practise integrating a product of logarithmic and algebraic functions.
Polynomial integration: The expression \(x^2+1\) must be integrated term by term.
Algebraic simplification: Dividing \(\frac{x^3}{3}+x\) by \(x\) correctly produces \(\frac{x^2}{3}+1\).
Stepwise presentation: Showing the selection of \(u\), \(dv\), \(du\), and \(v\) makes the application of integration by parts clear.

Significance for Competitive Entrance Examinations

JEE Main and Advanced: This is a standard application of integration by parts involving a logarithmic function and a polynomial expression.
Technique selection: Recognising that the logarithmic function should be differentiated simplifies the integral.
Polynomial manipulation: The problem reinforces the integration of polynomial expressions and simplification of rational algebraic terms.
Accuracy in multistep calculations: Correctly handling coefficients and negative signs is essential in longer integration problems.
Verification skills: Differentiation confirms the final antiderivative and helps identify errors in the product rule.

🚫 Common Mistakes

Common Mistakes

›

Incorrect choice of functions: Choose \(u=\log x\) and \(dv=(x^2+1)\,dx\).
Incorrect integration: Remember that \(\int(x^2+1)\,dx=\frac{x^3}{3}+x\).
Incorrect simplification: The expression \(\frac1x\left(\frac{x^3}{3}+x\right)\) simplifies to \(\frac{x^2}{3}+1\).
Coefficient error: The integral of \(\frac{x^2}{3}\) is \(\frac{x^3}{9}\), not \(\frac{x^3}{3}\).
Sign error: The entire remaining integral must be subtracted from the product \(uv\).
Domain restriction: For a real-valued logarithm, the integral is considered on \(x>0\).
Missing constant: Include \(C\) in the final answer.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 8 points

For \(\int(x^2+1)\log x\,dx\), integration by parts is the appropriate method.
According to ILATE, choose \(u=\log x\) and \(dv=(x^2+1)\,dx\).
The derivative of \(\log x\) is \(\frac1x\).
The integral of \(x^2+1\) is \(\frac{x^3}{3}+x\).
After applying integration by parts, simplify the remaining integrand before integrating.
Maintain the negative sign before the remaining integral.
The real-valued logarithmic function requires \(x>0\).
Always include the arbitrary constant of integration \(C\).

📘 Concept & Theory

Theory and Concept

›

The given integrand is the product of an exponential function and a sum of trigonometric functions. We can split the integral into two parts and evaluate them using integration by parts.

The integration by parts formula is:

\[ \boxed{\int u\,dv=uv-\int v\,du} \]

An important observation is that the derivatives of \(\sin x\) and \(\cos x\) are \(\cos x\) and \(-\sin x\), respectively. This creates a useful cancellation when the two integrals are combined.

🗺️ Solution Roadmap

Step-by-step Plan

›

Split the original integral into two integrals using the linearity property.
Evaluate the first integral \(\int e^x\sin x\,dx\) using integration by parts.
Evaluate the second integral \(\int e^x\cos x\,dx\) using integration by parts.
Combine the two results and simplify the expressions.
Verify the final answer by differentiation.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 24 steps

Given —
\[I=\int e^x(\sin x+\cos x)\,dx\]
Using the linearity property of integration:
\[\begin{aligned}I&=\int e^x\sin x\,dx+\int e^x\cos x\,dx\end{aligned}\]
Let:
\[I_1=\int e^x\sin x\,dx\]
and
\[I_2=\int e^x\cos x\,dx\]
Therefore:
\[I=I_1+I_2\]
Evaluate \(I_1=\int e^x\sin x\,dx\).
Applying integration by parts, choose:
\[u=\sin x,\quad dv=e^x\,dx\]
Differentiating \(u\):
\[du=\cos x\,dx\]
Integrating \(dv\):
\[v=e^x\]
Using the integration by parts formula:
\[\begin{aligned}I_1&=uv-\int v\,du\\ &=e^x\sin x-\int e^x\cos x\,dx\end{aligned}\]
Since:
\[I_2=\int e^x\cos x\,dx\]
We obtain:
\[\boxed{I_1=e^x\sin x-I_2}\tag{1}\]
Evaluate \(I_2=\int e^x\cos x\,dx\)
Again, applying integration by parts, choose:
\[u=\cos x,\quad dv=e^x\,dx\]
Differentiating \(u\):
\[du=-\sin x\,dx\]
Integrating \(dv\):
\[v=e^x\]
Applying integration by parts:
\[ \begin{aligned} I_2 &=uv-\int v\,du\\ &=e^x\cos x-\int e^x(-\sin x)\,dx\\ &=e^x\cos x+\int e^x\sin x\,dx \end{aligned} \]
Since:
\[I_1=\int e^x\sin x\,dx\]
We obtain:
\[\boxed{I_2=e^x\cos x+I_1}\tag{2}\]
Combine equations (1) and (2)
From equation (1):
\[I_1=e^x\sin x-I_2\]
Substituting equation (2):
\[ \begin{aligned} I_1 &=e^x\sin x-(e^x\cos x+I_1)\\ &=e^x\sin x-e^x\cos x-I_1 \end{aligned} \]
Taking \(I_1\) to the left-hand side:
\[ \begin{aligned} I_1+I_1 &=e^x\sin x-e^x\cos x\\ 2I_1 &=e^x(\sin x-\cos x) \end{aligned} \]
Therefore:
\[\boxed{I_1=\frac{e^x}{2}(\sin x-\cos x)}\]
Using equation (2):
\[ \begin{aligned} I_2 &=e^x\cos x+\frac{e^x}{2}(\sin x-\cos x)\\ &=\frac{e^x}{2}(\sin x+\cos x) \end{aligned} \]
Thus:
\[\boxed{I_2=\frac{e^x}{2}(\sin x+\cos x)}\]
Calculate the original integral
Recall:
\[I=I_1+I_2\]
Substituting both values:
\[ \begin{aligned} I &=\frac{e^x}{2}(\sin x-\cos x) +\frac{e^x}{2}(\sin x+\cos x)\\ &=\frac{e^x}{2} [\sin x-\cos x+\sin x+\cos x]\\ &=\frac{e^x}{2}(2\sin x)\\ &=e^x\sin x \end{aligned} \]
Hence, the required integral is:
\[\boxed{\displaystyle\int e^x(\sin x+\cos x)\,dx=e^x\sin x+C}\]

💡 Answer

Final Answer

›

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

🎯 Exam Significance

Exam Significance

›

Linearity of integration: The problem tests the ability to split an integral containing a sum into separate integrals.
Integration by parts: Students practise applying integration by parts to exponential and trigonometric functions.
Repeated integration: The problem demonstrates how two related integrals can be expressed in terms of each other and solved simultaneously.
Trigonometric differentiation: Correctly remembering the derivatives of sine and cosine is essential for handling signs.
Product rule: Recognising the derivative of \(e^x\sin x\) provides a quick way to verify the result.

Significance for Competitive Entrance Examinations

JEE Main and Advanced: This problem develops the ability to solve exponential-trigonometric integrals using integration by parts.
Simultaneous integral equations: Expressing \(I_1\) in terms of \(I_2\) and vice versa is a useful technique for integrals that recur after repeated integration by parts.
Pattern recognition: Identifying expressions that arise from the product rule can significantly reduce calculation time.
Sign accuracy: The negative derivative of \(\cos x\) is crucial in obtaining the correct equations.
Method efficiency: Comparing the direct product-rule observation with the longer integration-by-parts method improves problem-solving flexibility.

🚫 Common Mistakes

Common Mistakes

›

Incorrect differentiation: Remember that \(\frac{d}{dx}(\cos x)=-\sin x\).
Sign error in integration by parts: Carefully retain the negative sign in \(\int u\,dv=uv-\int v\,du\).
Incorrect equation substitution: From \(I_1=e^x\sin x-I_2\) and \(I_2=e^x\cos x+I_1\), obtain \(2I_1=e^x(\sin x-\cos x)\).
Incorrect combination: Do not cancel terms before correctly determining \(I_1\) and \(I_2\). The cosine terms cancel only when the two evaluated integrals are added.
Missing factor: When solving for \(I_1\) or \(I_2\), divide by \(2\).
Missing constant: Include \(C\) in the final answer.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 7 points

Use the linearity property to split the original integral into sine and cosine components.
Apply integration by parts to both component integrals.
Recognise that the resulting expressions contain the original component integrals.
Rearrange the equations algebraically to obtain \(I_1\) and \(I_2\).
When adding the two results, the cosine terms cancel and the sine terms combine.
The product rule provides a quick verification of the final result.
Always include the arbitrary constant of integration \(C\).

📘 Concept & Theory

Theory and Concept

›

This integral involves the product of an exponential function and a rational algebraic function. A useful technique is integration by parts, based on the identity:

\[ \int u\,dv=uv-\int v\,du \]

The algebraic factor can be differentiated, while the exponential factor can be integrated easily. However, it is important to check whether the resulting integral becomes simpler before proceeding.

🗺️ Solution Roadmap

Step-by-step Plan

›

Express the integrand as a product of \(e^x\) and \(\frac{x}{1+x^2}\).
Apply integration by parts.
Differentiate the rational function \(\frac{x}{1+x^2}\) using the quotient rule.
Substitute the derivative into the resulting integral.
Recognise the limitation of this transformation and use a suitable alternative method if necessary.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 8 steps

Given —
\[I=\int \frac{xe^x}{1+x^2}\,dx\]
Rewrite the integral as:
\[I=\int \left(\frac{x}{1+x^2}\right)e^x\,dx\]
Using integration by parts, take:
\[u=\frac{x}{1+x^2},\qquad dv=e^x\,dx\]
Therefore:
\[du=\frac{d}{dx}\left(\frac{x}{1+x^2}\right)dx\]

\[v=\int e^x\,dx=e^x\]
Applying the formula:
\[I=uv-\int v\,du\]

\[I=\frac{xe^x}{1+x^2}-\int e^x\frac{d}{dx}\left(\frac{x}{1+x^2}\right)dx\]
Differentiate the rational function
Using the quotient rule:
\[\frac{d}{dx}\left(\frac{x}{1+x^2}\right)=\frac{(1+x^2)\frac{d}{dx}(x)-x\frac{d}{dx}(1+x^2)}{(1+x^2)^2}\]
Since
\[\frac{d}{dx}(x)=1,\quad \frac{d}{dx}(1+x^2)=2x,\]
we obtain:
\[ \begin{aligned} \frac{d}{dx}\left(\frac{x}{1+x^2}\right) &=\frac{(1+x^2)-2x^2}{(1+x^2)^2}\\ &=\frac{1-x^2}{(1+x^2)^2} \end{aligned} \]
Substitute the derivative
Substituting into the integration-by-parts expression gives:
\[\boxed{I=\frac{xe^x}{1+x^2}-\int\frac{e^x(1-x^2)}{(1+x^2)^2}\,dx}\]

👁️ An Important Observation About This Method

Observation

›

The integration-by-parts step is mathematically correct, but it does not produce a simpler elementary integral. The remaining integral \[ \int\frac{e^x(1-x^2)}{(1+x^2)^2}\,dx \] is not an elementary integral that can be evaluated by the usual NCERT techniques.

In fact, the original integral itself does not have an elementary antiderivative. It can be represented using the exponential integral function, denoted by \(\operatorname{Ei}(z)\), which is an advanced special function rather than a standard elementary function.

🧮 Exact Representation Using a Special Function

Special Function

›

For completeness, the integral can be expressed as follows. First, use partial fractions:

\[\frac{x}{1+x^2}=\frac12\left(\frac{1}{x-i}+\frac{1}{x+i}\right)\]

Thus,

\[I=\frac12\int\frac{e^x}{x-i}\,dx+\frac12\int\frac{e^x}{x+i}\,dx\]

Using the identity

\[\int\frac{e^x}{x-a}\,dx=e^a\operatorname{Ei}(x-a)+C,\]

we obtain:

\[\boxed{I=\frac12\left[e^i\operatorname{Ei}(x-i)+e^{-i}\operatorname{Ei}(x+i)\right]+C}\]

Here, \(i=\sqrt{-1}\), and the two terms combine to give a real-valued antiderivative for real \(x\).

💡 Answer

Final Answer

›

\[\boxed{\int\frac{xe^x}{1+x^2}\,dx=\frac12\left[e^i\operatorname{Ei}(x-i)+e^{-i}\operatorname{Ei}(x+i)\right]+C}\]

🎯 Exam Significance

Exam Significance

›

The quotient rule must be applied carefully when differentiating rational functions.
Integration by parts is useful only when it simplifies the remaining integral or leads to a solvable relation.
Students should not assume that every integral presented in a question can be evaluated using elementary integration techniques.

8. Significance for Competitive Entrance Examinations

Recognising the structure of an integrand is essential when selecting an integration technique.
Careful algebraic simplification prevents errors in multi-step problems.
Advanced mathematics distinguishes elementary antiderivatives from expressions involving special functions such as \(\operatorname{Ei}(x)\).

🚫 Common Mistakes

Common Mistakes

›

Writing \(x(2x)=2x\) instead of \(2x^2\).
Incorrectly simplifying \(1+x^2-2x^2\) as \(1+x^2-2x\).
Assuming that integration by parts must always produce a simpler integral.
Presenting an unevaluated transformed integral as though it were a complete solution.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 4 points

Integration by parts gives
\[ I=\frac{xe^x}{1+x^2} -\int\frac{e^x(1-x^2)}{(1+x^2)^2}\,dx. \]
The correct derivative is
\[ \frac{d}{dx}\left(\frac{x}{1+x^2}\right) =\frac{1-x^2}{(1+x^2)^2}. \]
The resulting integral is not elementary.
The exact antiderivative requires the exponential integral function.

📘 Concept & Theory

Theory and Concept

›

The integrand is a product of an exponential function and a trigonometric rational expression. A suitable approach is to simplify the trigonometric expression before applying integration by parts.

We can use the identity:

\[ 1+\cos x=2\cos^2\frac{x}{2} \]

and

\[ 1+\sin x =\left(\sin\frac{x}{2}+\cos\frac{x}{2}\right)^2. \]

These identities give a useful half-angle representation of the quotient. We will also examine the derivative calculated in the supplied solution.

🗺️ Solution Roadmap

Step-by-step Plan

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Differentiate the trigonometric quotient using the quotient rule.
Correctly simplify the numerator using \(\sin^2x+\cos^2x=1\).
Apply integration by parts to the original integral.
Determine whether the resulting integral can be evaluated using elementary functions.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 6 steps

Given —
\[I=\int e^x\left(\frac{1+\sin x}{1+\cos x}\right)dx\]
Differentiate the trigonometric quotient
Let
\[f(x)=\frac{1+\sin x}{1+\cos x}\]
Using the quotient rule:
\[\begin{aligned}f'(x)&=\frac{(1+\cos x)\frac{d}{dx}(1+\sin x)-(1+\sin x)\frac{d}{dx}(1+\cos x)}{(1+\cos x)^2}\\ &=\frac{(1+\cos x)\cos x+(1+\sin x)\sin x}{(1+\cos x)^2}\\ &=\frac{\cos x+\cos^2x+\sin x+\sin^2x}{(1+\cos x)^2}\\ &=\frac{\sin x+\cos x+(\sin^2x+\cos^2x)}{(1+\cos x)^2}\\ &=\frac{1+\sin x+\cos x}{(1+\cos x)^2}.\end{aligned}\]
Apply integration by parts.
Take:
\[u=\frac{1+\sin x}{1+\cos x},\quad dv=e^x\,dx\]
Then:
\[du=\frac{1+\sin x+\cos x}{(1+\cos x)^2}\,dx,\quad v=e^x.\]
Using the integration-by-parts formula:
\[\begin{aligned}I&=uv-\int v\,du\\ &=e^x\left(\frac{1+\sin x}{1+\cos x}\right)-\int e^x\left(\frac{1+\sin x+\cos x}{(1+\cos x)^2}\right)dx.\end{aligned}\]
Thus,
\[\boxed{I=e^x\left(\frac{1+\sin x}{1+\cos x}\right)-\int\frac{e^x(1+\sin x+\cos x)}{(1+\cos x)^2}\,dx}\]

📝

Note

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Does Integration by Parts Complete the Solution?

No. Although the differentiation and integration-by-parts steps are correct, the remaining integral is not simpler in a way that leads to an elementary antiderivative. We must not treat the expression above as a completed evaluation.

In fact, the original integral does not have an elementary antiderivative. It can be expressed using special functions, but that is beyond the standard NCERT Class 12 integration syllabus.

Why the Integral Is Non-Elementary

Using the half-angle identity, \[ \frac{1+\sin x}{1+\cos x} = \frac{1+\tan(x/2)}{1-\tan(x/2)} \] wherever the expressions are defined.

This transformation does not eliminate the exponential-trigonometric interaction. A further substitution such as \(t=\tan(x/2)\) gives \[ e^{2\tan^{-1}t}\frac{1+t}{1-t}\frac{2}{1+t^2}\,dt, \] which is not an elementary rational-exponential integral.

Therefore, the usual NCERT techniques do not yield an elementary closed form for the given integrand.

💡 Answer

Final Answer

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\[ \boxed{ \begin{aligned} I={}&e^x\left(\frac{1+\sin x}{1+\cos x}\right)\\ &-\int\frac{e^x(1+\sin x+\cos x)} {(1+\cos x)^2}\,dx. \end{aligned} } \]

🎯 Exam Significance

Exam Significance

›

The quotient rule is essential for differentiating trigonometric rational expressions.
The identity \(\sin^2x+\cos^2x=1\) must be used carefully during simplification.
Integration by parts should be selected when it makes the remaining integral easier to evaluate.
A correct intermediate transformation does not necessarily constitute a complete evaluation.

Significance for Competitive Entrance Examinations

Questions involving products of exponential and trigonometric functions test the ability to identify useful substitutions and integration techniques.
Accurate differentiation of a quotient is important in multi-step calculus problems.
Recognising when a standard method fails is also an important mathematical skill.

🚫 Common Mistakes

Common Mistakes

›

Forgetting the negative sign in the derivative of \(\cos x\).
Incorrectly applying the quotient rule.
Simplifying \(\sin^2x+\cos^2x\) incorrectly.
Assuming that a correct integration-by-parts transformation must always lead to an elementary result.
Presenting an unevaluated integral as the final answer.

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 4 points

The supplied differentiation is correct.
The derivative of the quotient is:
\[ \frac{d}{dx}\left(\frac{1+\sin x}{1+\cos x}\right) = \frac{1+\sin x+\cos x}{(1+\cos x)^2}. \]
Integration by parts produces a valid intermediate expression, but does not complete the integration.
The given integral has no elementary antiderivative.

📘 Concept & Theory

Theory and Concept

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This integral involves an exponential function multiplied by a rational algebraic expression. The key is to recognise the derivative of a quotient involving \(e^x\).

Recall the product rule:

\[ \frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}. \]

We can use this rule in reverse to identify the integrand as the derivative of a product.

🗺️ Solution Roadmap

Step-by-step Plan

›

Split the given integral into two parts.
Apply integration by parts to the first integral.
Show that the remaining integral cancels with the second integral.
Obtain the final result and verify it by differentiation.

✏️ Solution

Complete Solution

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Step-by-step Solution · 8 steps

Given —
\[I=\int e^x\left(\frac{1}{x}-\frac{1}{x^2}\right)dx\]
Split the integrand
Using the linearity of integration:
\[I=\int\frac{e^x}{x}\,dx-\int\frac{e^x}{x^2}\,dx\]
Evaluate the first integral using integration by parts.
Let
\[u=\frac{1}{x},\qquad dv=e^x\,dx\]
Then:
\[du=-\frac{1}{x^2}\,dx,\quad v=e^x\]
Applying the integration-by-parts formula:
\[\int u\,dv=uv-\int v\,du\]
We obtain:
\[\begin{aligned}\int\frac{e^x}{x}\,dx&=\frac{e^x}{x}-\int e^x\left(-\frac{1}{x^2}\right)dx\\ &=\frac{e^x}{x}+\int\frac{e^x}{x^2}\,dx.\end{aligned}\]
Substitute the result into the original integral
We have:
\[I=\int\frac{e^x}{x}\,dx-\int\frac{e^x}{x^2}\,dx\]
Substituting the expression obtained in Step 2:
\[\begin{aligned}I&=\left[\frac{e^x}{x}+\int\frac{e^x}{x^2}\,dx\right]-\int\frac{e^x}{x^2}\,dx\\ &=\frac{e^x}{x}.\end{aligned}\]
Therefore, including the constant of integration:
\[\boxed{I=\frac{e^x}{x}+C}\]

💡 Answer

Final Answer

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🎯 Exam Significance

Exam Significance

›

Direct recognition of derivatives can significantly reduce calculation time.
The expression tests the relationship between differentiation of quotients and integration by parts.
Similar structures frequently appear in problems involving exponential functions and rational expressions.

🚫 Common Mistakes

Common Mistakes

›

Forgetting the negative sign when differentiating \(\frac{1}{x}\).
Incorrectly applying the integration-by-parts formula.
Failing to recognise that the remaining integrals cancel.
Omitting the constant of integration.

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 4 points

Integration by parts is applied using \(u=\frac{1}{x}\) and \(dv=e^x\,dx\).
The derivative of \(\frac{1}{x}\) is \(-\frac{1}{x^2}\), which produces a positive sign in the integration-by-parts result.
The two integrals involving \(\frac{e^x}{x^2}\) cancel each other.
The integrand can be recognised directly as the derivative of \(\frac{e^x}{x}\).

📘 Concept & Theory

Theory and Concept

›

This integral involves the product of an exponential function and a rational algebraic function. A useful strategy is to simplify the rational expression and look for a function whose derivative produces the given integrand.

We will use the product rule in reverse:

\[ \frac{d}{dx}[u(x)v(x)] =u'(x)v(x)+u(x)v'(x). \]

In particular, derivatives of expressions involving \(e^x\) and powers of \((x-1)\) can generate rational factors with the same denominator as the given integrand.

🗺️ Solution Roadmap

Step-by-step Plan

›

Rewrite the rational expression in terms of powers of \(x-1\).
Identify a suitable product involving \(e^x\).
Differentiate that product and compare it with the given integrand.
Verify the result by differentiation.

✏️ Solution

Complete Solution

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Step-by-step Solution · 6 steps

Given —
\[I=\int \frac{x-3}{(x-1)^3}\,e^x\,dx\]
Rewrite the numerator
Observe that:
\[x-3=(x-1)-2\]
Therefore,
\[ \begin{aligned} I &=\int\frac{(x-1)-2}{(x-1)^3}e^x\,dx\\ &=\int e^x\left[ \frac{1}{(x-1)^2} -\frac{2}{(x-1)^3} \right]dx. \end{aligned} \]
Identify a suitable derivative Consider the function:
\[F(x)=\frac{e^x}{(x-1)^2}\]
Using the product rule:
\[ \begin{aligned} F'(x) &=\frac{d}{dx}\left[e^x(x-1)^{-2}\right]\\ &=e^x(x-1)^{-2} +e^x\left[-2(x-1)^{-3}\right]\\ &=\frac{e^x}{(x-1)^2} -\frac{2e^x}{(x-1)^3}\\ &=e^x\left[ \frac{1}{(x-1)^2} -\frac{2}{(x-1)^3} \right]. \end{aligned} \]
Compare with the original integrand
The original integral is:
\[ I=\int e^x\left[ \frac{1}{(x-1)^2} -\frac{2}{(x-1)^3} \right]dx. \]
This is exactly the derivative of \(\frac{e^x}{(x-1)^2}\). Hence:
\[I=\frac{e^x}{(x-1)^2}+C.\]

💡 Answer

Final Answer

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

🎯 Exam Significance

Exam Significance

›

This problem illustrates the use of algebraic manipulation before integration.
It tests the application of the product rule in reverse to recognise an antiderivative.
Rewriting \(x-3\) as \((x-1)-2\) is the key simplification that reveals the required derivative.

Significance for Competitive Entrance Examinations

Recognising derivatives of products can reduce lengthy calculations to a few steps.
Similar patterns occur in integrals involving \(e^x\) multiplied by rational functions.
Identifying the appropriate algebraic rearrangement is often more important than applying a lengthy integration technique.

🚫 Common Mistakes

Common Mistakes

›

Expanding or simplifying the numerator incorrectly.
Forgetting the negative sign when differentiating \((x-1)^{-2}\).
Using the product rule without differentiating both factors.
Omitting the constant of integration.

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 4 points

Rewrite \(x-3\) as \((x-1)-2\).
The rational expression becomes:
\[\frac{x-3}{(x-1)^3}=\frac{1}{(x-1)^2}-\frac{2}{(x-1)^3}\]
The resulting integrand is the derivative of \(\frac{e^x}{(x-1)^2}\).
Always verify an indefinite integral by differentiation.

📘 Concept & Theory

Theory and Concept

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This integral involves the product of an exponential function and a trigonometric function. It can be evaluated using integration by parts twice.

The integration-by-parts formula is:

\[ \int u\,dv=uv-\int v\,du. \]

When integration by parts is applied twice to products of exponential and trigonometric functions, the original integral reappears. We can then form an algebraic equation in the integral and solve for it.

🗺️ Solution Roadmap

Step-by-step Plan

›

Apply integration by parts to the given integral, taking the exponential function as the part to integrate.
Evaluate the resulting integral using integration by parts again.
Substitute the second result into the first equation.
Collect the original integral terms and solve for \(I\).
Verify the result by differentiation.

✏️ Solution

Complete Solution

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Step-by-step Solution · 13 steps

Given —
\[I=\int e^{2x}\sin x\,dx\]
Apply integration by parts for the first time
Choose:
\[u=\sin x,\quad dv=e^{2x}\,dx\]
Then:
\[du=\cos x\,dx,\quad v=\int e^{2x}\,dx=\frac{e^{2x}}{2}\]
Using integration by parts:
\[ \begin{aligned} I &=uv-\int v\,du\\ &=\frac{e^{2x}}{2}\sin x -\frac12\int e^{2x}\cos x\,dx \end{aligned} \]
Let:
\[J=\int e^{2x}\cos x\,dx\]
Therefore:
\[\boxed{I=\frac{e^{2x}\sin x}{2}-\frac{J}{2}}\tag{1}\]
Evaluate \(J\) using integration by parts again
We have:
\[J=\int e^{2x}\cos x\,dx\]
Choose:
\[u=\cos x,\quad dv=e^{2x}\,dx\]
Then:
\[du=-\sin x\,dx,\quad v=\frac{e^{2x}}{2}\]
Applying integration by parts:
\[ \begin{aligned} J &=uv-\int v\,du\\ &=\frac{e^{2x}}{2}\cos x -\int\frac{e^{2x}}{2}(-\sin x)\,dx\\ &=\frac{e^{2x}\cos x}{2} +\frac12\int e^{2x}\sin x\,dx \end{aligned} \]
Since \(I=\int e^{2x}\sin x\,dx\), we get:
\[\boxed{J=\frac{e^{2x}\cos x}{2}+\frac{I}{2}}\tag{2}\]
Substitute the value of \(J\)
Substituting equation (2) into equation (1):
\[ \begin{aligned} I &=\frac{e^{2x}\sin x}{2} -\frac12\left( \frac{e^{2x}\cos x}{2}+\frac{I}{2} \right)\\ &=\frac{e^{2x}\sin x}{2} -\frac{e^{2x}\cos x}{4} -\frac{I}{4}. \end{aligned} \]
Solve for \(I\)
Collecting the terms containing \(I\):
\[I+\frac{I}{4}=\frac{e^{2x}\sin x}{2}-\frac{e^{2x}\cos x}{4}\]

\[\frac{5I}{4}=\frac{e^{2x}}{4}(2\sin x-\cos x)\]
Multiplying both sides by \(\frac45\):
\[I=\frac{e^{2x}}{5}(2\sin x-\cos x)+C\]

💡 Answer

Final Answer

›

\[\boxed{\int e^{2x}\sin x\,dx=\frac{e^{2x}}{5}(2\sin x-\cos x)+C}\]

🎯 Exam Significance

Exam Significance

›

This is a standard application of repeated integration by parts.
It demonstrates how the original integral reappears after applying integration by parts twice.
Students must carefully handle the negative sign arising from the derivative of \(\cos x\).
Forming and solving an algebraic equation in the original integral is an important technique for evaluating such expressions.

6. Significance for Competitive Entrance Examinations

This integral tests the ability to recognise cyclic integration-by-parts patterns.
The same technique applies to integrals of the forms:
\[ \int e^{ax}\sin bx\,dx \quad\text{and}\quad \int e^{ax}\cos bx\,dx. \]
Efficient manipulation of the resulting algebraic equations helps reduce calculation time in timed examinations.

🚫 Common Mistakes

Common Mistakes

›

Writing \(\int e^{2x}\,dx=e^{2x}\) instead of \(\frac{e^{2x}}2\).
Forgetting that \(\frac{d}{dx}(\cos x)=-\sin x\).
Incorrectly handling the negative sign in the second integration-by-parts step.
Failing to substitute the second integral back into the first equation.
Omitting the constant of integration in the final result.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 4 points

Apply integration by parts twice.
Introduce a second integral:
\[ J=\int e^{2x}\cos x\,dx. \]
After the second application, the original integral \(I\) reappears.
Collecting the \(I\)-terms gives the required result.

📘 Concept & Theory

Theory and Concept

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This integral involves an inverse trigonometric function with a rational expression as its argument. The key is to recognise the double-angle identity:

\[ \sin 2\theta=\frac{2\tan\theta}{1+\tan^2\theta}. \]

By putting \(x=\tan\theta\), the argument of the inverse sine can be transformed into \(\sin 2\theta\). However, an important consideration is that the principal value of \(\sin^{-1}(\sin 2\theta)\) is not always equal to \(2\theta\). We must account for the range of the inverse sine function.

We will first simplify the inverse trigonometric expression and then apply integration by parts.

🗺️ Solution Roadmap

Step-by-step Plan

›

Use the double-angle identity to simplify the argument.
Account for the principal-value range of the inverse sine function.
Apply integration by parts.
Evaluate the resulting integral using a suitable substitution.
State the result on the appropriate intervals and verify it by differentiation.

✏️ Solution

Complete Solution

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Step-by-step Solution · 22 steps

Given —
\[I=\int\sin^{-1}\left(\frac{2x}{1+x^2}\right)dx\]
Recall:
\[ \sin 2\theta=\frac{2\tan\theta}{1+\tan^2\theta}. \]
Taking \(x=\tan\theta\), we obtain:
\[ \frac{2x}{1+x^2}=\sin 2\theta. \]
Therefore,
\[\sin^{-1}\left(\frac{2x}{1+x^2}\right)=\sin^{-1}(\sin 2\theta)\]
Since the principal range of \(\sin^{-1}y\) is
\[ -\frac{\pi}{2}\leq\sin^{-1}y\leq\frac{\pi}{2}, \]
we cannot simply replace \(\sin^{-1}(\sin 2\theta)\) by \(2\theta\) for every real \(x\).
Determine the principal-value expression.
For \(x\geq 0\), take \(\theta=\tan^{-1}x\), so \(0\leq\theta<\frac{\pi}{2}\). The principal-value identity gives:
\[ \sin^{-1}(\sin 2\theta)= \begin{cases} 2\theta,&0\leq\theta\leq\frac{\pi}{4},\\ \pi-2\theta,&\frac{\pi}{4}\leq\theta<\frac{\pi}{2} \end{cases} \]
Since \(\theta=\tan^{-1}x\), this becomes:
\[ \sin^{-1}\left(\frac{2x}{1+x^2}\right) = \begin{cases} 2\tan^{-1}x,&0\leq x\leq1,\\ \pi-2\tan^{-1}x,&x\geq1. \end{cases} \]
For negative \(x\), the expression is odd. Thus, the complete real-valued identity is:
\[ \boxed{ \sin^{-1}\left(\frac{2x}{1+x^2}\right) = \begin{cases} -\pi-2\tan^{-1}x,&x\leq-1,\\ 2\tan^{-1}x,&-1\leq x\leq1,\\ \pi-2\tan^{-1}x,&x\geq1. \end{cases} } \]
Evaluate the integral for \(-1\leq x\leq1\)
In this interval:
\[I=\int 2\tan^{-1}x\,dx\]
Using integration by parts, take:
\[u=2\tan^{-1}x,\quad dv=dx\]
Then:
\[du=\frac{2}{1+x^2}dx,\quad v=x\]
Applying the formula:
\[ \begin{aligned} I &=uv-\int v\,du\\ &=2x\tan^{-1}x -2\int\frac{x}{1+x^2}dx \end{aligned} \]
Evaluate the remaining integral
Put:
\[t=1+x^2\]
Then:
\[dt=2x\,dx\]
Therefore:
\[ \begin{aligned} 2\int\frac{x}{1+x^2}dx &=\int\frac{dt}{t}\\ &=\ln|t|\\ &=\ln(1+x^2) \end{aligned} \]
Since \(1+x^2>0\), the absolute-value signs are unnecessary.
Hence, for \(-1\leq x\leq1\):
\[\boxed{I=2x\tan^{-1}x-\ln(1+x^2)+C}\]
Complete Real-Domain Result
The original integrand is defined for every real \(x\), but its principal-value expression changes at \(x=-1\) and \(x=1\). Consequently, the antiderivative must be written piecewise.
Case I: For \(x\leq-1\)
\[I=\int(-\pi-2\tan^{-1}x)\,dx\]
Using
\[\int\tan^{-1}x\,dx=x\tan^{-1}x-\frac12\ln(1+x^2),\]
we obtain:
\[\boxed{I=-\pi x-2x\tan^{-1}x+\ln(1+x^2)+C_1}\]
Case II: For \(-1\leq x\leq1\)
\[\boxed{I=2x\tan^{-1}x-\ln(1+x^2)+C_2}\]
Case III: For \(x\geq1\)
\[I=\int(\pi-2\tan^{-1}x)\,dx\]
Therefore:
\[\boxed{I=\pi x-2x\tan^{-1}x+\ln(1+x^2)+C_3}\]
If a single antiderivative is required across the entire real line, the constants must be chosen to make the pieces agree continuously at \(x=-1\) and \(x=1\).

💡 Answer

Final Answer

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For the commonly intended textbook interval \(-1\leq x\leq1\):

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

Important mathematical note: This boxed expression is valid on \([-1,1]\). Outside that interval, the principal-value identity changes, so the corresponding piecewise antiderivatives given above must be used.

🎯 Exam Significance

Exam Significance

›

This problem combines inverse trigonometric identities with integration by parts.
It tests understanding of the principal-value range of inverse trigonometric functions.
The substitution \(t=1+x^2\) simplifies the remaining logarithmic integral.
Students should distinguish between an identity that holds on a restricted interval and one that holds for all real values.

Significance for Competitive Entrance Examinations

Principal-value restrictions are important in inverse trigonometric problems.
The identity involving \(\sin 2\theta\) provides an efficient route to simplify the integrand.
Piecewise analysis is necessary whenever an inverse trigonometric function is composed with a periodic trigonometric function.
Checking the domain and range prevents incorrect simplifications that can lead to wrong answers.

🚫 Common Mistakes

Common Mistakes

›

Assuming \(\sin^{-1}(\sin\theta)=\theta\) for every real \(\theta\).
Ignoring the principal range of the inverse sine function.
Using \(2\tan^{-1}x\) as the inverse-sine expression when \(|x|>1\).
Forgetting the factor of \(2\) while applying integration by parts.
Writing \(\ln|1+x^2|\) unnecessarily; \(1+x^2\) is always positive.

🔑 Key Takeaways

Key Takeaways

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Key Takeaways · 4 points

Use the identity:
\[ \frac{2x}{1+x^2} =\sin(2\tan^{-1}x). \]
The simplification to \(2\tan^{-1}x\) is valid only for \(-1\leq x\leq1\).
Integration by parts gives:
\[ \int 2\tan^{-1}x\,dx =2x\tan^{-1}x-\ln(1+x^2)+C. \]
For the full real domain, the answer must be expressed piecewise.

📘 Concept & Theory

Theory and Concept

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This integral can be evaluated using the substitution method. The integrand contains an exponential function whose exponent is \(x^3\), along with the factor \(x^2\).

Recall the standard integration formula:

\[ \int e^u\,du=e^u+C. \]

The derivative of the exponent \(x^3\) is \(3x^2\). Therefore, the factor \(x^2\,dx\) is directly related to the differential of the exponent.

🗺️ Solution Roadmap

Step-by-step Plan

›

Substitute the exponent \(x^3\) with a new variable \(u\).
Differentiate the substitution to obtain \(du\).
Rewrite the integral in terms of \(u\).
Integrate and substitute back \(u=x^3\).
Compare the result with the given options.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 9 steps

Apply substitution
Let:
\[u=x^3\]
Differentiating both sides with respect to \(x\):
\[\frac{du}{dx}=3x^2\]
Therefore:
\[du=3x^2\,dx\]
Rearranging:
\[x^2\,dx=\frac{1}{3}du\]
Substitute into the integral
The given integral is:
\[I=\int x^2e^{x^3}\,dx\]
Using \(u=x^3\):
\[ \begin{aligned} I &=\int e^u\left(\frac13du\right)\\ &=\frac13\int e^u\,du. \end{aligned} \]
Integrate
Using the standard formula:
\[\int e^u\,du=e^u+C,\]
we obtain:
\[I=\frac13e^u+C\]
Substitute back
Since \(u=x^3\):
\[\boxed{I=\frac13e^{x^3}+C}\]

💡 Answer

Final Answer

›

Correct option: (A)

\[\boxed{\int x^2e^{x^3}\,dx=\frac13e^{x^3}+C}\]

🎯 Exam Significance

Exam Significance

›

This question tests the substitution method of integration.
Students must identify the relationship between the derivative of the exponent and the remaining algebraic factor.
It reinforces the chain rule in reverse, which is fundamental to evaluating composite exponential integrals.

7. Significance for Competitive Entrance Examinations

Such questions test quick recognition of standard substitution patterns.
Understanding the coefficient introduced by differentiation helps eliminate incorrect options efficiently.
The same approach applies to integrals of the form:
\[ \int [f(x)]^n e^{[f(x)]^{n+1}}f'(x)\,dx \]
whenever the remaining factors match the derivative of the exponent.

🚫 Common Mistakes

Common Mistakes

›

Forgetting the coefficient \(3\) when differentiating \(x^3\).
Choosing \(u=x^2\) instead of the exponent \(x^3\).
Confusing the exponent in the answer with the power of the algebraic factor.
Omitting the constant of integration.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 4 points

Choose the exponent as the substitution: \(u=x^3\).
The differential is \(du=3x^2\,dx\).
The coefficient \(\frac13\) appears because of the derivative of \(x^3\).
Always verify the selected option by differentiation.

📘 Concept & Theory

Theory and Concept

›

This integral involves the product of an exponential function and a trigonometric expression. The most effective approach is to recognise the derivative of a product.

Recall the product rule:

\[ \frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}. \]

In particular, the derivative of \(\sec x\) is \(\sec x\tan x\). This suggests that the given integrand may be the derivative of \(e^x\sec x\).

🗺️ Solution Roadmap

Step-by-step Plan

›

Expand the trigonometric factor in the integrand.
Identify the derivative of the product \(e^x\sec x\).
Match the derivative with the given integrand.
Verify the result and compare the options.

✏️ Solution

Complete Solution

›

Step-by-step Solution · 4 steps

Given —
\[I=\int e^x\sec x(1+\tan x)\,dx\]
Expanding the bracket
\[I=\int e^x(\sec x+\sec x\tan x)\,dx\]
Recognise the derivative of a product
Consider:
\[F(x)=e^x\sec x\]
Using the product rule:
\[ \begin{aligned} F'(x) &=\frac{d}{dx}(e^x)\sec x +e^x\frac{d}{dx}(\sec x)\ &=e^x\sec x+e^x\sec x\tan x\ &=e^x\sec x(1+\tan x). \end{aligned} \]
The derivative is exactly equal to the given integrand.
Evaluate the integral
Therefore:
\[ \begin{aligned} I &=\int\frac{d}{dx}(e^x\sec x)\,dx\\ &=e^x\sec x+C. \end{aligned} \]

💡 Answer

Final Answer

›

Correct option: (B)

\[\boxed{\int e^x\sec x(1+\tan x)\,dx=e^x\sec x+C}\]

🎯 Exam Significance

Exam Significance

›

This problem tests the application of the product rule in reverse.
Students must remember the standard derivative:
\[ \frac{d}{dx}(\sec x)=\sec x\tan x. \]
Recognising a complete derivative is often more efficient than applying integration by parts.

Significance for Competitive Entrance Examinations

Direct recognition of derivatives is an important time-saving technique in objective questions.
The problem tests knowledge of exponential and trigonometric differentiation together.
Similar patterns can be solved by identifying derivatives of products such as \(e^x\sin x\), \(e^x\cos x\), and \(e^x\sec x\).

🚫 Common Mistakes

Common Mistakes

›

Forgetting the derivative of the exponential factor \(e^x\).
Using \(\frac{d}{dx}(\sec x)=\sec^2x\), which is incorrect.
Confusing the derivatives of \(\sec x\) and \(\tan x\).
Ignoring the constant of integration.

🔑 Key Takeaways

Key Takeaways

›

Key Takeaways · 4 points

Expand the trigonometric expression when necessary.
Use the product rule in reverse to identify the antiderivative.
The derivative of \(e^x\sec x\) produces the exact integrand.
Differentiate the selected option to verify its correctness.

Integrals

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

8. Significance for Competitive Entrance Examinations

Does Integration by Parts Complete the Solution?

Why the Integral Is Non-Elementary

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

6. Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

7. Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Chapter Complete!

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INTEGRALS — Learning Resources

Frequently Asked Questions

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Integrals

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

8. Significance for Competitive Entrance Examinations

Does Integration by Parts Complete the Solution?

Why the Integral Is Non-Elementary

Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

6. Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

7. Significance for Competitive Entrance Examinations

Significance for Competitive Entrance Examinations

Chapter Complete!

Recent posts

INTEGRALS — Learning Resources

Frequently Asked Questions

Q 01 What is the correct answer to NCERT Class 12 Mathematics Integrals Exercise 7.6 Question 24?

Q 02 How do you evaluate ? e^x sec x (1 + tan x) dx?

Q 03 Which differentiation rule is used to solve this integration problem?

Q 04 What is the derivative of sec x?

Q 05 What is the derivative of e^x sec x?

Q 06 Why is Option (B) correct?

Q 07 Can this integral be solved using integration by parts?

Q 08 Why is the constant of integration C included in the answer?

Q 09 What is the main concept tested in Exercise 7.6 Question 24?

Q 10 How is this question useful for CBSE Board and JEE examinations?

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