$\smallint \,\,\frac{{{{\sin }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}\;\,\,dx$ is equal to

$\log \,\,({\sin ^{ - 1}}x) + c$

$\log \,\,\left( {\sqrt {1 - {x^2}} } \right) + c$

$\smallint \,\,\frac{{\sin \,\,x\,\,\cos \,\,x}}{{\sqrt {1\,\, - \,\,{{\sin }^4}\,\,x} }}\,\,dx$ is equal to

$\frac{1}{2}\,\,{\cos ^{ - 1}}\,\,({\sin ^2}\,\,x) + c$

${\tan ^{ - 1}}\,\,\left( {{{\sin }^2}\,\,x} \right) + c$

${\tan ^{ - 1}}\,\,\left( {2\,\,{{\sin }^2}\,\,x} \right) + c$

The antiderivative of $\frac{1}{\sin ^2 x+\tan ^2 x}$ is

$(-1 / 2)\left[\tan x+(1 / \sqrt{2}) \tan ^{-1}(1 / \sqrt{2}) \tan x\right]+C$

The value of $\smallint \,\,\frac{{{{\cos }^3}\,\,x\,\, + \,\,{{\cos }^5}\,\,x}}{{{{\sin }^2}\,\,x\,\, + \,\,{{\sin }^4}\,\,x}}\,\,dx$, is

$\sin \,\,x\,\, - \,\,6\,\,{\tan ^{ - 1}}\,\,\left( {\sin \,\,x} \right) + C$

$\sin \,\,x\,\, - \,\,2\,\,{\left( {\sin \,\,x} \right)^{ - 1}} + \,\,C$

$\sin \,\,x\,\, - \,\,2\,\,{\left( {\sin \,\,x} \right)^{ - 1}}\,\, + \,\,5\,\,{\tan ^{ - 1}}\,\,\left( {\sin \,\,x} \right)\,\, + \,\,C$

If $\int\left\{\log (\log x)+\frac{1}{(\log x)^2}\right\} d x=x\{f(x)-g(x)\}+C,$ then

$f(x)=\log x ; g(x)=\frac{1}{\log x}$

$f(x)=\frac{1}{\log x} ;(x)=\log (\log x)$

$f(x)=\frac{1}{x \log x} ; g(x)=\frac{1}{\log x}$

$\smallint \,\,\frac{{m{x^{m + 2n - 1}}\,\, - \,\,n{x^{n - 1}}}}{{{x^{2m + 2n}}\,\, + \,\,2{x^{m + n}}\,\, + \,\,1}}\;\,\,dx$ is equal to

$\frac{{{x^m}}}{{{x^{m + n}}\,\, + \,\,1}}\,\, + \,\,c$

$\frac{{{x^{m + n}}\,\, - \,\,1}}{{{x^{m + n}}\,\, + \,\,1}}\,\, + \,\,c$

$\smallint \,\,\frac{{{e^x}\left( {1 + \,\sin \,\,x} \right)}}{{1 + \cos \,\,x}}\,\,dx$ is equal to

${e^x}\,\,\left( {\frac{{1 + \sin \,\,x}}{{1 - \cos \,\,x}}} \right)\,\, + \,\,c$

$c\,\, - \,\,{e^x}\,\,\cot \left( {\frac{x}{2}} \right)$

If $\int \tan ^7 x \mathrm{~d} x=f(x)+C$ then

$f(x)$ is a polynomial of degree 8 in $\tan x$

$f(x)$ is a polynomial of degree 5 in $\tan x$

$f(x)$ is a polynomial of degree 6 in $\tan x$

$\smallint \,\,\frac{{{\rm{cosec}}\,\,x}}{{{{\cos }^2}\,\,\left( {1\,\, + \,\,\log \,\,\tan \,\,\frac{x}{2}} \right)}}\,\,dx$

${\sin ^2}\,\,\left[ {1 + \,\,\log \,\,\tan \,\,\frac{x}{2}} \right]\,\, + \,\,c$

${\sec ^2}\,\,\left[ {1 + \,\,\log \,\,\tan \,\,\frac{x}{2}} \right]\,\, + \,\,c$

$\smallint \frac{{{e^{2x}} - \,\,2{e^x}}}{{{e^{2x}} + 1}}\,\,dx$ is equal to

$\log \,\,\left( {{e^{2x}} + 1} \right) - \,\,{\tan ^{ - 1}}\,\,\left( {{e^x}} \right) + C$

$\frac{1}{2}\,\,\log \,\,\left( {{e^{2x}} + \,\,1} \right) - {\tan ^{ - 1}}\,\,\left( {{e^x}} \right) + C$

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JEE • MATHEMATICS • INDEFINITE INTEGRATION

Indefinite integration practice · Easy

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 241-270 of 746 questions

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241 Easy

241 of 746

Application +4 | -1

\int \,\,\frac{{{{\sin }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}\;\,\,dx

is equal to

View answer and explanation

Correct option: B

\frac{1}{2}\,\,{\left( {{{\sin }^{ - 1}}x} \right)^2} + c

Option B (

\frac{1}{2}\,{\left( {{{\sin }^{ - 1}}x} \right)^2} + c

) is correct because it results from a straightforward u-substitution with

u = \sin^{-1}x

242 Easy

242 of 746

Analyzing +4 | -1

\int \,\,\frac{{2{x^2}\,\, + \,\,3}}{{\left( {{x^2} - 1} \right)\,\,\left( {{x^2} + 4} \right)}}\,\,dx = a\,\,\log \,\,\left( {\frac{{x + 1}}{{x - 1}}} \right)\,\, + \,\,b\,\,{\tan ^{ - 1}}\,\,\frac{x}{2}

, then

\left( {a,\,\;b} \right)

View answer and explanation

Correct option: A

\left( { - 1/2,\;\,\,1/2} \right)

Option A ((-1/2, 1/2)) is correct because by performing partial fraction decomposition and integrating, we find the coefficients a and b to be -1/2 and 1/2, respectively.

243 Easy

243 of 746

Application +4 | -1

\int \,\,\frac{{\sin \,\,x\,\,\cos \,\,x}}{{\sqrt {1\,\, - \,\,{{\sin }^4}\,\,x} }}\,\,dx

is equal to

View answer and explanation

Correct option: A

\frac{1}{2}\,\,{\sin ^{ - 1}}\,\,\left( {{{\sin }^2}\,\,x} \right) + c

Option A (

\frac{1}{2}\,{\sin ^{ - 1}}\,\,\left( {{{\sin }^2}\,\,x} \right) + c

) is correct because after substitution, the integral simplifies to a standard arcsin form.

244 Easy

244 of 746

Application +4 | -1

I=\int \frac{\left(x+x^{\frac{2}{3}}+x^{\frac{1}{6}}\right)}{x\left(1+x^{\frac{1}{3}}\right)} d x

is equal to

View answer and explanation

Correct option: A

\frac{3}{2} x^{\frac{2}{3}}+6 \tan ^{-1}\left(x^{\frac{1}{6}}\right)+c

Option A (

\frac{3}{2} x^{\frac{2}{3}}+6 \tan ^{-1}\left(x^{\frac{1}{6}}\right)+c

) is correct because it results from the integration after a suitable substitution to simplify the expression.

245 Easy

245 of 746

Application +4 | -1

The value of

\int \frac{\sin x+\cos x}{3+\sin 2 x} d x

, is

View answer and explanation

Correct option: A

\frac{1}{4} \log \left(\frac{2+\sin x-\cos x}{2-\sin x+\cos x}\right)+C

Option A is correct because the integral of the given function can be solved by manipulating the denominator and using substitution, leading to a logarithmic form.

246 Easy

246 of 746

Application +4 | -1

\int \frac{\left(\tan ^{-1} x\right)^3}{1+x^2} d x

is equal to

View answer and explanation

Correct option: B

\frac{\left(\tan ^{-1} x\right)^4}{4}+C

Option B is correct because the integral simplifies to (tan^(-1)(x))⁴ / 4 + C after recognizing the derivative of tan^(-1)(x).

247 Easy

247 of 746

Application +4 | -1

The antiderivative of

\frac{1}{\sin ^2 x+\tan ^2 x}

View answer and explanation

Correct option: B

(-1 / 2)\left[\cot x+(1 / \sqrt{2}) \tan ^{-1}((1 / \sqrt{2}) \tan x)\right]+C

Option B is the correct antiderivative, obtained by simplifying the integrand, substituting \

t = \tan x

and integrating the resulting expression.

248 Easy

248 of 746

Application +4 | -1

Let

f

be a function such that

f\left( 1 \right) = 4

and

f'\left( x \right) \ge 2

for

1 \le x < 4

. How small can

f\left( 4 \right)

possibly be?

View answer and explanation

Correct option: D

The correct answer is D (10) because it represents the smallest possible value of f(4) based on the given conditions and the lower bound of the derivative.

249 Easy

249 of 746

Analyzing +4 | -1

Let

f(x)=\int \frac{x^2}{\left(1+x^2\right)\left(1+\sqrt{1+x^2}\right)} d x

and

f(0)=0

. Then

f(1)

is equal to

View answer and explanation

Correct option: B

\log _e(1+\sqrt{2})-\frac{\pi}{4}

Option B (

\log _e(1+\sqrt{2})-\frac{\pi}{4}

) is correct because it is the result of evaluating the definite integral with the given initial condition.

250 Easy

250 of 746

Analyzing +4 | -1

The value of

\int \,\,\frac{{{{\cos }^3}\,\,x\,\, + \,\,{{\cos }^5}\,\,x}}{{{{\sin }^2}\,\,x\,\, + \,\,{{\sin }^4}\,\,x}}\,\,dx

, is

View answer and explanation

Correct option: C

\sin \,\,x\,\, - \,\,2\,\,{\left( {\sin \,\,x} \right)^{ - 1}}\,\, - \,\,6\,\,{\tan ^{ - 1}}\,\,\left( {\sin \,\,x} \right) + C

Option C is correct because the integral simplifies to

\sin x - 2(\sin x)^{-1} - 6\tan^{-1}(\sin x) + C

after substitution and partial fraction decomposition.

251 Easy

251 of 746

Application +4 | -1

\int \frac{\cos 2 \theta}{(\sin \theta+\cos \theta)^2} d \theta

is equal to

View answer and explanation

Correct option: B

\log |\sin \theta+\cos \theta|+c

The correct answer is B (3) because by using the given conditions and trigonometric identities, we can derive that cot(alpha)cot(gamma) = 3.

252 Easy

252 of 746

Application +4 | -1

\int\left\{\log (\log x)+\frac{1}{(\log x)^2}\right\} d x=x\{f(x)-g(x)\}+C,

then

View answer and explanation

Correct option: A

f(x)=\log (\log x) ; g(x)=\frac{1}{\log x}

Option A (

f(x)=\log (\log x) ; g(x)=\frac{1}{\log x}

) is correct because after integration by parts and substitution, the given integral matches the form

x\{f(x)-g(x)\} + C

253 Easy

253 of 746

Analyzing +4 | -1

\int \,\,\frac{{\cos \,\,4x\,\, + \,\,1}}{{\cot \,\,x\,\, - \,\,\tan \,\,x}}\,\,dx = k\,\,\cos \,\,4x + \,\,c

, then

View answer and explanation

Correct option: B

k = - 1/8

Option B (

k = - 1/8

) is correct because after simplifying and integrating the given expression, the coefficient of

\cos 4x

is -1/8.

254 Easy

254 of 746

Analyzing +4 | -1

\int \,\,\frac{{m{x^{m + 2n - 1}}\,\, - \,\,n{x^{n - 1}}}}{{{x^{2m + 2n}}\,\, + \,\,2{x^{m + n}}\,\, + \,\,1}}\;\,\,dx

is equal to

View answer and explanation

Correct option: D

- \,\,\frac{{{x^n}}}{{{x^{m + n}}\,\, + \,\,1}}\,\, + \,\,c

Option D (

- \,\frac{{{x^n}}}{{{{x^{m + n}}\\ + \\,1}}}\\ + \\,c

) is correct because it is the result of the integration after appropriate substitution and simplification.

255 Easy

255 of 746

Application +4 | -1

\int \,\,\frac{{{e^x}\left( {1 + \,\sin \,\,x} \right)}}{{1 + \cos \,\,x}}\,\,dx

is equal to

View answer and explanation

Correct option: A

{e^x}\,\,\tan \,\,\left( {\frac{x}{2}} \right) + c

Option A is correct because the integral can be simplified and solved using the properties of exponential functions and trigonometric identities.

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SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs DIFFERENTIAL EQUATIONS 977 MCQs

256 Easy

256 of 746

Application +4 | -1

The value of

\int \frac{\sin \alpha}{\sqrt{1+\cos \alpha}} \mathrm{d} \alpha

View answer and explanation

Correct option: B

-2 \sqrt{2} \cos (\alpha / 2)+C

Option B (

-2 \sqrt{2} \cos (\alpha / 2)+C

) is correct because the integral of the given expression simplifies to this result after substitution.

257 Easy

257 of 746

Application +4 | -1

\int a^{3 x+3} d x=

View answer and explanation

Correct option: B

\frac{a^{3 x+3}}{3 \log a}+c

Option B is correct because the integral of

a^{3x+3}

\frac{a^{3x+3}}{3 \log a} + c

after applying the substitution method.

258 Easy

258 of 746

Application +4 | -1

\int \frac{3 \sin x+2 \cos x}{3 \cos x+2 \sin x} d x=a x+b \log |3 \cos x+2 \sin x|+C

, then

View answer and explanation

Correct option: B

a=\frac{12}{13}, b=-\frac{5}{13}

Option B (

a=\frac{12}{13}, b=-\frac{5}{13}

) is correct because the values of a and b are obtained by comparing coefficients after expressing the numerator in terms of the derivative and the original denominator.

259 Easy

259 of 746

Application +4 | -1

\int \tan ^7 x \mathrm{~d} x=f(x)+C

then

View answer and explanation

Correct option: C

f(x)=\frac{1}{6} \tan ^6 x-\frac{1}{4} \tan ^4 x+\frac{1}{2} \tan ^2 x+\log |\cos x|+C

Option C is correct because the integral of tan⁷(x) can be found by using substitution and polynomial division.

260 Easy

260 of 746

Application +4 | -1

\int \,\,\frac{{\mathrm{cosec}\,\,x}}{{{{\cos }^2}\,\,\left( {1\,\, + \,\,\log \,\,\tan \,\,\frac{x}{2}} \right)}}\,\,dx

View answer and explanation

Correct option: B

\tan \,\,\left[ {1 + \,\,\log \,\,\tan \,\,\frac{x}{2}} \right]\,\, + \,\,c

Option B is correct because after the trigonometric simplification and substitution, the integral is a standard form, resulting in tan[1 + log tan(x/2)] + c.

261 Easy

261 of 746

Application +4 | -1

\int \frac{1-x^2}{\left(1+x^2\right) \sqrt{1+x^4}} d x

is equal to

View answer and explanation

Correct option: B

\frac{1}{\sqrt{2}} \sin ^{-1}\left\{\frac{\sqrt{2} x}{x^2+1}\right\}+C

Option B is correct. By dividing the numerator and denominator by

x^2

and using a substitution, the integral can be simplified to a standard inverse sine integral.

262 Easy

262 of 746

Application +4 | -1

\int \frac{d x}{e^x-1}=

View answer and explanation

Correct option: A

\ln \left(1-e^{-x}\right)+c

Option A is correct because the integral of

\frac{1}{e^x-1}

can be solved by rewriting the integrand as

\frac{e^{-x}}{1-e^{-x}}

and using substitution.

263 Easy

263 of 746

Analyzing +4 | -1

\int \frac{x^3-1}{x^3+x} d x

is equal to

View answer and explanation

Correct option: B

x-\log x+\frac{1}{2} \log \left(x^2+1\right)-\tan ^{-1} x+C

Option B (

x-\log x+\frac{1}{2} \log \left(x^2+1\right)-\tan ^{-1} x+C

) is correct because it is the result of integrating the given expression after partial fraction decomposition.

264 Easy

264 of 746

Application +4 | -1

\int \frac{x^{5 / 2}}{\sqrt{1+x^7}} d x

, is

View answer and explanation

Correct option: A

\frac{2}{7} \log \left|x^{7 / 2}+\sqrt{1+x^7}\right|+C

Option A is correct because the integral is solved using a substitution method, resulting in the given logarithmic expression.

265 Easy

265 of 746

Analyzing +4 | -1

\int \tan x \tan 2 x \tan 3 x \,d x

is equal to

View answer and explanation

Correct option: B

\frac{1}{3} \log |\sec 3 x|-\frac{1}{2} \log |\sec 2 x|-\log |\sec x|+C

Option B is correct because using the identity

\tan 3x = \tan(2x+x)

and simplifying the integral, we get

\frac{1}{3} \log |\sec 3 x|-\frac{1}{2} \log |\sec 2 x|-\log |\sec x|+C

266 Easy

266 of 746

Application +4 | -1

\int \frac{{{e^{2x}} - \,\,2{e^x}}}{{{e^{2x}} + 1}}\,\,dx

is equal to

View answer and explanation

Correct option: C

\frac{1}{2}\,\,\log \,\,\left( {{e^{2x}} + 1} \right) - 2\,\,{\tan ^{ - 1}}\,\,\left( {{e^x}} \right) + C

Option C (

\frac{{1}}{{2}}\,\log \,\left( {{e^{2x}} + 1} \right) - 2\,{\tan ^{ - 1}}\,\left( {{e^x}} \right) + C

) is correct because it is the result of direct integration after separating the original integral into two simpler integrals.

267 Easy

267 of 746

Application +4 | -1

\int \log x d x=

View answer and explanation

Correct option: B

x \log x-x+c

Option B (

x \log x-x+c

) is correct because it is the result of integrating

\log x

using integration by parts.

268 Easy

268 of 746

Application +4 | -1

\int \frac{\sin x+\cos x}{\sin (x-\alpha)} d x

is equal to

View answer and explanation

Correct option: A

(\cos \alpha-\sin \alpha)(x-\alpha)+(\cos \alpha+\sin \alpha) \log |\sin (x-\alpha)|+C

Option A is correct because after substituting

t = x - \alpha

and simplifying, the integral evaluates to the expression in option A.

269 Easy

269 of 746

Application +4 | -1

\int\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2 d x

is equal to

View answer and explanation

Correct option: D

\frac{x^2}{2}+\log x-2 x+c

Option D (

\frac{x^2}{2}+\log x-2 x+c

) is correct because it is the result of expanding the square and integrating each term.

270 Easy

270 of 746

Application +4 | -1

\int {{e^x}\left( {4{x^2} + 8x + 3} \right)dx}

is equal to

View answer and explanation

Correct option: C

\left( {4{x^2} + 3} \right){e^x} + k

Option C (

\left( {4{x^2} + 3} \right){e^x} + k

) is correct because the integral can be solved by recognizing a pattern of the form

\int e^x [f(x) + f'(x)] dx = e^x f(x) + C

Revision notes and key takeaways

Review INDEFINITE INTEGRATION in your textbook or class notes, then use this set to check your understanding.

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