Evaluate: $\int x\tan^{-1}x\,dx$

Evaluate: ∫ xtan⁻¹xdx

3.3 · Q109

\int x\tan^{-1}x\,dx

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Concept understanding — Integration by Parts

Integration by Parts

The idea: reverse the product rule

Some integrands are a product of two very different functions — $x e^x$ , $x\cos x$ , $\log x$ , $x\sin^{-1}x$ — where substitution gets you nowhere. Integration by parts is the tool for these. It comes straight from reversing the product rule for differentiation.

Starting from $\dfrac{d}{dx}(uv)=u\,v' + u'\,v$ and integrating both sides gives the working formula:

\int u\,\frac{dv}{dx}\,dx = uv - \int v\,\frac{du}{dx}\,dx.

In words: integral of (first $\times$ derivative-of-second) $=$ first $\times$ integral-of-second $-$ integral of (derivative-of-first $\times$ integral-of-second).

Choosing $u$ : the ILATE rule

The whole game is picking which factor is $u$ (to differentiate) and which is $dv$ (to integrate). Pick $u$ by ILATE — the first type that appears:

Inverse trig ( $\sin^{-1}x$ ), Logarithmic ( $\log x$ ), Algebraic ( $x^2$ ), Trigonometric ( $\sin x$ ), Exponential ( $e^x$ ).

Whatever comes first in ILATE becomes $u$ ; the rest is $dv$ . This makes the new integral $\int v\,du$ simpler than the one you started with.

Worked idea

For $\int x e^x\,dx$ : algebraic before exponential, so $u=x$ , $dv=e^x dx$ . Then $du=dx$ , $v=e^x$ :

\int x e^x\,dx = x e^x - \int e^x\,dx = x e^x - e^x + C = e^x(x-1)+C.

Tip

A single log or a single inverse-trig function ( $\int\log x\,dx$ , $\int\sin^{-1}x\,dx$ ) is still "by parts" — take the other factor as $1$ . And for the special form $\int e^x\big(f(x)+f'(x)\big)dx$ , the answer is simply $e^x f(x)+C$ .

Note

If applying the formula gives you back a multiple of the original integral (as with $\int e^x\sin x\,dx$ ), don't panic — solve for the integral algebraically.

Integration by Parts is one of the most tested methods in the NCERT Class 12 Mathematics chapter on Integrals, and "integration by parts formula ILATE rule" along with "integration by parts class 12 important questions" are among the top searches for students preparing for CBSE board exams and JEE Main calculus. The same ILATE-based technique extends naturally into JEE Advanced integral calculus problems built on this NCERT Class 12 foundation.

$u=\tan^{-1}x$ , $dv=x\,dx\Rightarrow v=\dfrac{x^2}2$ , $\dfrac{du}{dx}=\dfrac1{1+x^2}$ .

\int x\tan^{-1}x\,dx=\dfrac{x^2}2\tan^{-1}x-\int\dfrac{x^2}{2(1+x^2)}dx

Write $\dfrac{x^2}{1+x^2}=1-\dfrac1{1+x^2}$ , so

\int\dfrac{x^2}{2(1+x^2)}dx=\dfrac12\left(x-\tan^{-1}x\right)

\int x\tan^{-1}x\,dx=\dfrac{x^2}2\tan^{-1}x-\dfrac{x}2+\dfrac12\tan^{-1}x+c=\dfrac{x^2+1}{2}\tan^{-1}x-\dfrac{x}2+c

✓Final answer

\dfrac{x^2+1}{2}\tan^{-1}x-\dfrac{x}{2}+c

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 $v=x$ :

$\int\log x\,dx=x\log x-\int x\cdot\tfrac{1}{x}\,dx=x\log x-\int 1\,dx=x\log x-x+k$ .

✓Final answer
$x\log x - x + k$
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 ( $u=t,\ dv=\cos t\,dt$ ): $=t\sin t-\int\sin t\,dt=t\sin t+\cos t$ .

Hence the answer is $2(t\sin t+\cos t)+k=2(\sqrt{x}\sin\sqrt{x}+\cos\sqrt{x})+k$ .

✓Final answer
$2(\sqrt{x}\sin\sqrt{x}+\cos\sqrt{x})+k$
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

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

Therefore the integral equals $e^x\tan^{-1}x+k$ .

✓Final answer
$e^x\tan^{-1}x+k$
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 $du=dx,\ v=e^x$ :

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

Evaluate from $0$ to $1$ : $(1-1)e^1-(0-1)e^0=0-(-1)=1$ .

✓Final answer
$1$
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

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

which exactly matches the given integrand.

Therefore

$\int e^x(\sin x-\cos x)\,dx=e^x\cdot(-\cos x)+c=-e^x\cos x+c$

✓Final answer
$-e^x\cos x+c$
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 $f'(x) = \dfrac{\sec x\tan x}{\sec x} = \tan x$ .

So the integrand $e^x(\log\sec x + \tan x) = e^x[f(x)+f'(x)]$ , and by the standard result:

$\int e^x(\log\sec x+\tan x)\,dx = e^x \log\sec x + C$ .

✓Final answer
$e^x(\log\sec x)+C$
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).

Take $f(x) = -\cos x$ , so $f'(x) = \sin x$ . Then $f + f' = -\cos x + \sin x = \sin x - \cos x$ , exactly the integrand. Hence $\int e^x(\sin x - \cos x)\,dx = e^x f(x) = -e^x\cos x + C$ .

✓Final answer
$-e^x \cos x$
CBSE 2025Set ANNUAL1 mark
$\displaystyle\int x \cdot e^x\,dx$
›Reveal solutionSolution

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

$\int x\,e^x\,dx = x\,e^x - \int e^x\,dx = x\,e^x - e^x + c = e^x(x-1)+c$

✓Final answer
$e^x(x-1) + c$
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 $u=\log x$ , $dv=dx$ :

$\int\log x\,dx=x\log x-\int x\cdot\frac1x\,dx=x\log x-x.$

Therefore

$\int\log x^2\,dx=2\int\log x\,dx=2(x\log x-x)+k.$

✓Final answer
$2(x\log x-x)+k$
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 $du=dx$ , $v=-\cos x$ .

$\int x\sin x\,dx=uv-\int v\,du=-x\cos x-\int(-\cos x)\,dx=-x\cos x+\int\cos x\,dx$

$=-x\cos x+\sin x+C$

✓Final answer
$-x\cos x+\sin x+C$
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

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

And $\displaystyle\frac12\int x\,dx=\frac12\cdot\frac{x^2}{2}=\frac{x^2}{4}$ , so

$I=\frac{x^2}{2}\log 2x-\frac{x^2}{4}+C.$

✓Final answer
$\displaystyle\int x\log 2x\,dx=\frac{x^2}{2}\log 2x-\frac{x^2}{4}+C$
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):

$\frac{d}{dx}\big(e^x\sec x\big)=e^x\sec x+e^x\sec x\tan x=e^x\sec x(1+\tan x),$

which is the original integrand. This confirms option (b).

✓Final answer
$e^x\sec x+c$

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