Evaluate: $\displaystyle\int_0^1 x\tan^{-1}x\,dx$.

> [!TLDR] > This integral is solved using integration by parts, choosing $\tan^{-1}x$ as the function to differentiate. The final value is $\boxed{\frac{\pi}{4} - \frac{1}{2}}$. When faced with an integral involving a product of two different types of functions, such as an algebraic function ($x$) and an inverse trigonometric function ($\tan^{-1}x$), the most common and effective technique is **integration by parts**. The core idea behind integration by parts is to transform a difficult integral into a potentially simpler one. The formula for definite integrals is: > [!FORMULA] > $$ \int_a^b u\,dv = [uv]_a^b - \int_a^b v\,du $$ The crucial step is choosing which part of the integrand becomes $u$ and which becomes $dv$. A helpful mnemonic for this choice is **LIATE**: * **L**ogarithmic functions * **I**nverse trigonometric functions * **A**lgebraic functions * **T**rigonometric functions * **E**xponential functions The function that appears earlier in the LIATE list is generally chosen as $u$, because its derivative tends to simplify the expression, or its integral is more complex than its derivative. In our problem, $\int_0^1 x\tan^{-1}x\,dx$: * $\tan^{-1}x$ is an **I**nverse trigonometric function. * $x$ is an **A**lgebraic function. According to LIATE, we should choose $u = \tan^{-1}x$ and $dv = x\,dx$. This choice is strategic because the derivative of $\tan^{-1}x$ is a simple rational function, and the integral of $x$ is straightforward. Let's proceed with the evaluation: 1. **Identify $u$ and $dv$**: Based on the LIATE rule, we set: $u = \tan^{-1}x$ $dv = x\,dx$ 2. **Calculate $du$ and $v$**: Differentiating $u$: $du = \frac{d}{dx}(\tan^{-1}x)\,dx = \frac{1}{1+x^2}\,dx$ Integrating $dv$: $v = \int x\,dx = \frac{x^2}{2}$ 3. **Apply the integration by parts formula**: Substitute $u$, $v$, $du$, and $dv$ into the formula $\int_0^1 u\,dv = [uv]_0^1 - \int_0^1 v\,du$: $$ \int_0^1 x\tan^{-1}x\,dx = \left

Evaluate: ∫₀¹ xtan⁻¹xdx.

Question

\displaystyle\int_0^1 x\tan^{-1}x\,dx

CBSECBSE Class XII Board 2026Subjective· 3mImportance★★★★★

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This integral is solved using integration by parts, choosing $\tan^{-1}x$ as the function to differentiate. The final value is $\boxed{\frac{\pi}{4} - \frac{1}{2}}$ .

When faced with an integral involving a product of two different types of functions, such as an algebraic function ( $x$ ) and an inverse trigonometric function ( $\tan^{-1}x$ ), the most common and effective technique is integration by parts.

The core idea behind integration by parts is to transform a difficult integral into a potentially simpler one. The formula for definite integrals is:

\int_a^b u\,dv = [uv]_a^b - \int_a^b v\,du

The crucial step is choosing which part of the integrand becomes $u$ and which becomes $dv$ . A helpful mnemonic for this choice is LIATE:

Logarithmic functions
Inverse trigonometric functions
Algebraic functions
Trigonometric functions
Exponential functions

The function that appears earlier in the LIATE list is generally chosen as $u$ , because its derivative tends to simplify the expression, or its integral is more complex than its derivative.

In our problem, $\int_0^1 x\tan^{-1}x\,dx$ :

$\tan^{-1}x$ is an Inverse trigonometric function.
$x$ is an Algebraic function.

According to LIATE, we should choose $u = \tan^{-1}x$ and $dv = x\,dx$ . This choice is strategic because the derivative of $\tan^{-1}x$ is a simple rational function, and the integral of $x$ is straightforward.

Let's proceed with the evaluation:

Identify $u$ and $dv$ :

Based on the LIATE rule, we set:

$u = \tan^{-1}x$

$dv = x\,dx$
Calculate $du$ and $v$ :

Differentiating $u$ :

$du = \frac{d}{dx}(\tan^{-1}x)\,dx = \frac{1}{1+x^2}\,dx$

Integrating $dv$ :

$v = \int x\,dx = \frac{x^2}{2}$
Apply the integration by parts formula:

Substitute $u$ , $v$ , $du$ , and $dv$ into the formula $\int_0^1 u\,dv = [uv]_0^1 - \int_0^1 v\,du$ :

\int_0^1 x\tan^{-1}x\,dx = \left[\frac{x^2}{2}\tan^{-1}x\right]_0^1 - \int_0^1 \frac{x^2}{2} \cdot \frac{1}{1+x^2}\,dx

Evaluate the first term:

\left[\frac{x^2}{2}\tan^{-1}x\right]_0^1 = \left(\frac{1^2}{2}\tan^{-1}(1)\right) - \left(\frac{0^2}{2}\tan^{-1}(0)\right)

We know that $\tan^{-1}(1) = \frac{\pi}{4}$ and $\tan^{-1}(0) = 0$.

= \left(\frac{1}{2} \cdot \frac{\pi}{4}\right) - \left(0 \cdot 0\right) = \frac{\pi}{8}

Evaluate the remaining integral: The integral we need to solve is $\int_0^1 \frac{x^2}{2(1+x^2)}\,dx$ . We can factor out the constant $\frac{1}{2}$ : $\frac{1}{2}\int_0^1 \frac{x^2}{1+x^2}\,dx$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 40 on this concept.

CBSE 2026Set A1 markMCQ
$\int\log x\,dx =$
(a) $\frac{1}{x} + k$
(b) $x\log x + k$
(c) $x\log x - x + k$
(d) $x\log x + x + k$
›Reveal solutionSolution

Integrate by parts: $\int\log x\,dx=x\log x-x+k$ .

Take $u=\log x$ and $dv=dx$ , so $du=\tfrac{1}{x}\,dx$ and $v=x$ :

…

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CBSE 2026Set A1 markMCQ
$\int\cos\sqrt{x}\,dx =$
(a) $\sin\sqrt{x} + \cos\sqrt{x} + k$
(b) $\frac{1}{2}(\sqrt{x}\sin\sqrt{x} - \cos\sqrt{x}) + k$
(c) $2(\sqrt{x}\sin\sqrt{x} + \cos\sqrt{x}) + k$
(d) $\sin\sqrt{x} + k$
›Reveal solutionSolution

Put $t=\sqrt{x}$ ; the integral becomes $2\int t\cos t\,dt=2(t\sin t+\cos t)+k$ .

Let $t=\sqrt{x}$ , so $x=t^2$ and $dx=2t\,dt$ . Then

$\int\cos\sqrt{x}\,dx=\int\cos t\,(2t\,dt)=2\int t\cos t\,dt$ .

Integrate $\int t\cos t\,dt$ by parts ( $u=t,\ dv=\cos t\,dt$ ): $=t\sin t-\int\sin t\,dt=t\sin t+\cos t$ .

…

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CBSE 2026Set A1 markMCQ
$\int e^x\left(\tan^{-1}x + \frac{1}{1+x^2}\right)dx =$
(a) $e^x\tan^{-1}x + k$
(b) $e^x \cdot \frac{1}{1+x^2} + k$
(c) $e^x + k$
(d) $\tan^{-1}x + k$
›Reveal solutionSolution

Recognise $\int e^x[f(x)+f'(x)]\,dx=e^x f(x)+k$ ; here $f(x)=\tan^{-1}x$ .

Note $\dfrac{d}{dx}\tan^{-1}x=\dfrac{1}{1+x^2}$ . So the integrand is $e^x\big[\tan^{-1}x+(\tan^{-1}x)'\big]$ , which matches the standard pattern

…

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CBSE 2026Set A1 markMCQ
$\int_0^1 x e^x\,dx =$
(a) $1$
(b) $0$
(c) $2$
(d) $-1$
›Reveal solutionSolution

By parts: $\int_0^1 xe^x\,dx=[(x-1)e^x]_0^1=1$ .

Take $u=x,\ dv=e^x\,dx$ , so $du=dx,\ v=e^x$ :

$\int x e^x\,dx=xe^x-\int e^x\,dx=xe^x-e^x=(x-1)e^x$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$\int e^x(\sin x-\cos x)\,dx$
(a) $-e^x\cos x+c$
(b) $e^x\sin x+c$
(c) $-e^x\sec x+c$
(d) $e^x\operatorname{cosec} x+c$
›Reveal solutionSolution

Recognising the standard form $\int e^x[f(x)+f'(x)]\,dx=e^xf(x)+c$ gives $-e^x\cos x+c$ .

There is a standard integration result:

$\int e^x\big[f(x)+f'(x)\big]\,dx=e^x f(x)+c$

Compare the integrand $\sin x-\cos x$ with $f(x)+f'(x)$ . Try $f(x)=-\cos x$ ; then $f'(x)=\sin x$ , so

$f(x)+f'(x)=-\cos x+\sin x=\sin x-\cos x$ …

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CBSE 2026Set ANNUAL1 markMCQ
$\int e^x (\log \sec x + \tan x)\, dx =$
(a) $e^x + C$
(b) $e^x \tan x + C$
(c) $e^x(\log \sec x) + C$
(d) None of these
›Reveal solutionSolution

This is of the standard form $\int e^x[f(x)+f'(x)]\,dx = e^x f(x)+C$ .

Let $f(x) = \log\sec x$ . Then $f'(x) = \dfrac{\sec x\tan x}{\sec x} = \tan x$ .

…

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CBSE 2025Set X11 markMCQ
$\int e^x (\sin x - \cos x)\, dx$
(a) $-e^x \cos x$
(b) $e^x \cos x$
(c) $e^x \sin x$
(d) $e^x \sin^2 x$
›Reveal solutionSolution

Integral of the form $\int e^x(f+f')\,dx = e^x f$ — correct option (a). …

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CBSE 2025Set ANNUAL1 mark
$\displaystyle\int x \cdot e^x\,dx$
›Reveal solutionSolution

Apply integration by parts with $u=x$ , $dv=e^x dx$ .

…

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CBSE 2025Set E1 markMCQ
$\int \log x^2\,dx =$
(a) $\frac{1}{x^2} + k$
(b) $\frac{2}{x} + k$
(c) $x\log x - x + k$
(d) $2(x\log x - x) + k$
›Reveal solutionSolution

Bring down the power, then integrate $\log x$ by parts; result $2(x\log x-x)+k$ .

First $\log x^2=2\log x$ . Now integrate $\int\log x\,dx$ by parts with $u=\log x$ , $dv=dx$ : …

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CBSE 2025Set ANNUAL1 mark
$\displaystyle\int x\sin x\,dx$
›Reveal solutionSolution

Use integration by parts (ILATE): take $u=x$ (algebraic) and $dv=\sin x\,dx$ .

Let $u=x$ , $dv=\sin x\,dx$ , so $du=dx$ , $v=-\cos x$ .

…

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CBSE 2025Set ANNUAL1 mark
$\displaystyle\int x\log 2x\, dx$
›Reveal solutionSolution

Use integration by parts, taking the logarithm as the first function.

Let $I=\displaystyle\int x\log 2x\,dx$ . Apply $\int u\,dv=uv-\int v\,du$ with

$u=\log 2x,\qquad dv=x\,dx.$

Then

$du=\frac{1}{2x}\cdot 2\,dx=\frac1x\,dx,\qquad v=\frac{x^2}{2}.$

Therefore

$I=\log 2x\cdot\frac{x^2}{2}-\int\frac{x^2}{2}\cdot\frac1x\,dx=\frac{x^2}{2}\log 2x-\frac12\int x\,dx.$ …

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CBSE 2025Set ANNUAL1 markMCQ
$\displaystyle\int e^x\sec x(1+\tan x)\, dx$
(a) $e^x\cos x + c$
(b) $e^x\sec x + c$
(c) $e^x\sin x + c$
(d) $e^x\tan x + c$
›Reveal solutionSolution

Recognise the form $\int e^x\{f(x)+f'(x)\}\,dx=e^xf(x)+c$ .

Expand the integrand:

$e^x\sec x(1+\tan x)=e^x\big(\sec x+\sec x\tan x\big).$

Let $f(x)=\sec x$ . Then

$f'(x)=\sec x\tan x.$

So the integrand is exactly $e^x\{f(x)+f'(x)\}$ , and by the standard result

$\int e^x\{f(x)+f'(x)\}\,dx=e^xf(x)+c=e^x\sec x+c.$

Verification (differentiate the answer): …

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Q.Evaluate: $\displaystyle\int_0^1 x\tan^{-1}x\,dx$ .

Concept understanding — Integration by Parts

Integration by Parts

The idea: reverse the product rule

Choosing $u$ : the ILATE rule

Worked idea

Q.Evaluate: ∫01xtan⁡−1x dx\displaystyle\int_0^1 x\tan^{-1}x\,dx∫01​xtan−1xdx.

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Q.Evaluate: $\displaystyle\int_0^1 x\tan^{-1}x\,dx$ .

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