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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 1-30 of 746 questions
1 Easy
1 of 746
Application +4 | -1
∫sin⁡5xcos⁡4xdx=\int \sin ^5 x \cos ^4 x d x=
View answer and explanation
Correct option: A
−15cos⁡5x+27cos⁡7x−19cos⁡9x+c-\frac{1}{5} \cos ^5 x+\frac{2}{7} \cos ^7 x-\frac{1}{9} \cos ^9 x+c
Option A is correct because the integral is solved by substituting cos⁡x=t\cos x = t and simplifying the resulting expression.
2 Easy
2 of 746
Analyzing +4 | -1
If ∫1−7cos⁡2xsin⁡7xcos⁡2xdx=f(x)sin⁡7x+C\int \frac{1-7 \cos ^2 x}{\sin ^7 x \cos ^2 x} d x=\frac{f(x)}{\sin ^7 x}+C is equal to
View answer and explanation
Correct option: C
tan⁡x\tan x
Option D (cot⁡x\cot x) is correct because by rewriting the integrand and integrating, f(x)f(x) is found to be tan⁡x\tan x, which is equivalent to sin⁡xcos⁡x\frac{\sin x}{\cos x} and simplifies to tan⁡x\tan x.
3 Easy
3 of 746
Application +4 | -1
∫x7(1−x2)5dx\int \frac{x^7}{\left(1-x^2\right)^5} d x is equal to:
View answer and explanation
Correct option: D
18⋅x8(1−x2)4\frac{1}{8} \cdot \frac{x^8}{\left(1-x^2\right)^4}
Option D, 18⋅x8(1−x2)4\frac{{1}}{{8}} \cdot \frac{{x^8}}{{(1-x^2)^4}}, is correct because using trigonometric substitution x=sin⁡θx = \sin \theta simplifies the integral, leading to this result.
4 Easy
4 of 746
Application +4 | -1
∫ ax3+  bx2+cx4   dx\int \,\frac{{a{x^3} + \,\,b{x^2} + c}}{{{x^4}}}\;\,dx equals
View answer and explanation
Correct option: C
a  log⁡  x  −  bx  −  c3x3  +  ka\,\,\log \,\,x\,\, - \,\,\frac{b}{x}\,\, - \,\,\frac{c}{{3{x^3}}}\,\, + \,\,k
Option C, a  log⁡ x  −  bx  −  c3x3  +  ka\,\,\log \,x\,\, - \,\,\frac{b}{x}\,\, - \,\,\frac{c}{{3{x^3}}}\,\, + \,\,k, is correct because it is the result of directly integrating each term after dividing by x4x^4.
5 Easy
5 of 746
Application +4 | -1
∫x2+2x4+4dx\int \frac{x^2+2}{x^4+4} d x equals
View answer and explanation
Correct option: C
12Tan⁡−1(x2−22x)+c\frac{1}{2} \operatorname{Tan}^{-1}\left(\frac{x^2-2}{2 x}\right)+c
Option C (12Tan⁡−1(x2−22x)+c\frac{1}{2} \operatorname{Tan}^{-1}\left(\frac{x^2-2}{2 x}\right)+c) is correct because after algebraic manipulation and substitution, the integral evaluates to this expression.
6 Easy
6 of 746
Application +4 | -1
∫x(sec⁡2x−1sec⁡2x+1)dx=\int x\left(\frac{\sec 2 x-1}{\sec 2 x+1}\right) d x=
View answer and explanation
Correct option: A
xtan⁡x−log⁡∣sec⁡x∣−x22+cx \tan x-\log |\sec x|-\frac{x^2}{2}+c
Option A is correct because the integral is solved by simplifying the trigonometric expression, then using integration by parts and standard integrals.
7 Easy
7 of 746
Application +4 | -1
∫sin⁡−1  x  dx\int {\sin ^{ - 1}}\,\,x\,\,dx is equal to
View answer and explanation
Correct option: B
x  sin⁡−1  x+  1−x2  +  cx\,\,{\sin ^{ - 1}}\,\,x + \,\,\sqrt {1 - {x^2}} \,\, + \,\,c
Option B (x  sin⁡−1 x+ 1−x2  +  cx\,\,{\sin ^{ - 1}}\,x + \,\sqrt {1 - {x^2}} \,\, + \,\,c) is correct because it is the result of integrating arcsin(x) by parts.
8 Easy
8 of 746
Application +4 | -1
∫dxex+1−2e−x=\int \frac{d x}{e^x+1-2 e^{-x}}=
View answer and explanation
Correct option: C
13log⁡(ex−1)−13log⁡(ex+2)+c\frac{1}{3} \log \left(e^x-1\right)-\frac{1}{3} \log \left(e^x+2\right)+c
Option A is correct because the integral is solved by multiplying the numerator and denominator by exe^x, then substituting t=ext = e^x, and using partial fraction decomposition.
9 Easy
9 of 746
Application +4 | -1
If ∫dxx2+2x+1=Alog⁡∣x+1∣+C\int \frac{{dx}}{\sqrt{{x}^2+2 {x}+1}}={A} \log |{x}+1|+{C} for x<−1{x} < -1 then AA is
View answer and explanation
Correct option: C
-1
Option C (-1) is correct because the integral simplifies to −log⁡∣x+1∣+C-\log |x+1| + C for x<−1x < -1, thus A = -1.
10 Easy
10 of 746
Application +4 | -1
∫xsin⁡xsec⁡3xdx=\int x \sin x \sec ^3 x d x=
View answer and explanation
Correct option: B
12[xsec⁡2x−tan⁡x]+c\frac{1}{2}\left[x \sec ^2 x-\tan x\right]+c
Option B (12[xsec⁡2x−tan⁡x]+c\frac{{1}}{{2}}\left[x \sec ^2 x-\tan x\right]+c) is correct because it is the result of integrating the given expression using substitution and integration by parts.
11 Easy
11 of 746
Application +4 | -1
If ∫x4a6+x6dx=g(x)+C\int \sqrt{\frac{x^4}{a^6+x^6}} d x=g(x)+C, then g(x)=g(x)=
View answer and explanation
Correct option: C
13log⁡∣x3+a6+x6∣\frac{1}{3} \log \left|x^3+\sqrt{a^6+x^6}\right|
Option C, 13log⁡∣x3+a6+x6∣\frac{1}{3} \log \left|x^3+\sqrt{a^6+x^6}\right|, is the correct antiderivative obtained by substitution and using the integral formula for ∫1u2+a2du\int \frac{1}{\sqrt{u^2+a^2}} du.
12 Easy
12 of 746
Application +4 | -1
∫cos⁡5xdx=\int \cos ^5 x d x=
View answer and explanation
Correct option: A
sin⁡x−23sin⁡3x+15sin⁡5x+c\sin x-\frac{2}{3} \sin ^3 x+\frac{1}{5} \sin ^5 x+c
Option A is correct because the integral can be solved by rewriting the integrand and using substitution, resulting in sin⁡x−23sin⁡3x+15sin⁡5x+c\sin x-\frac{2}{3} \sin ^3 x+\frac{1}{5} \sin ^5 x+c.
13 Easy
13 of 746
Application +4 | -1
∫e3log⁡x(x4+1)−1dx\int e^{3 \log x}\left(x^4+1\right)^{-1} d x is equal
View answer and explanation
Correct option: B
14log⁡(x4+1)+C\frac{1}{4} \log \left(x^4+1\right)+C
Option B (14log⁡(x4+1)+C\frac{1}{4} \log \left(x^4+1\right)+C) is correct because after simplifying the integral, the result matches this option.
14 Easy
14 of 746
Analyzing +4 | -1
The integral ∫3x13+2x11(2x4+3x2+1)4dx\int {\frac{{3{x^{13}} + 2{x^{11}}}}{{{{\left( {2{x^4} + 3{x^2} + 1} \right)}^4}}}dx} is equal to (where C is a constant of integration)
View answer and explanation
Correct option: B
x126(2x4+3x2+1)3+C\frac{{{x^{12}}}}{{6{{\left( {2{x^4} + 3{x^2} + 1} \right)}^3}}} + C
Option B (x126(2x4+3x2+1)3+C\frac{{{x^{12}}}}{{6{{\left( {2{x^4} + 3{x^2} + 1} \right)}^3}}} + C) is correct because after dividing the numerator and denominator by x16{x^{16}} and substituting, the integral can be easily evaluated.
15 Easy
15 of 746
Application +4 | -1
The indefinite integral of (12sin⁡x+5cos⁡x)−1(12 \sin x+5 \cos x)^{-1} is, for any arbitrary constant
View answer and explanation
Correct option: A
1/13log⁡∣tan⁡(x2+12tan⁡−1512)∣+c1 / 13 \log \left|\tan \left(\frac{\mathrm{x}}{2}+\frac{1}{2} \tan ^{-1} \frac{5}{12}\right)\right|+\mathrm{c}
Option A (1/13log⁡∣tan⁡(x2+12tan⁡−1512)∣+c1 / 13 \log \left|\tan \left(\frac{\mathrm{x}}{2}+\frac{1}{2} \tan ^{-1} \frac{5}{12}\right)\right|+\mathrm{c}) is correct because after applying trigonometric identities and integrating, the result matches this option.

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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
16 Easy
16 of 746
Analyzing +4 | -1
The value of ∫(1+x)x(1+xex)2dx\int \frac{(1+x)}{x\left(1+x e^x\right)^2} d x, is equal to
View answer and explanation
Correct option: B
log⁡∣xex1+xex∣+11+xex+C\log \left|\frac{x e^x}{1+x e^x}\right|+\frac{1}{1+x e^x}+C
Option B (log⁡∣xex1+xex∣+11+xex+C\log \left|\frac{x e^x}{1+x e^x}\right|+\frac{1}{1+x e^x}+C) is correct because the integral simplifies to this logarithmic form after appropriate substitutions and partial fraction decomposition.
17 Easy
17 of 746
Application +4 | -1
∫1+sin⁡  x4 dx\int \sqrt {1 + \sin \,\,\frac{x}{4}} \,dx is equal to
View answer and explanation
Correct option: A
8  (sin⁡  x8  −  cos⁡  x8)+c8\,\,\left( {\sin \,\,\frac{x}{8}\,\, - \,\,\cos \,\,\frac{x}{8}} \right) + c
Option A (8  (sin⁡  x8  −  cos⁡  x8)+c8\,\,\left( {\sin \,\,\frac{x}{8}\,\, - \,\,\cos \,\,\frac{x}{8}} \right) + c) is correct because it is the result of simplifying the integrand using trigonometric identities and integrating.
18 Easy
18 of 746
Application +4 | -1
∫sec⁡xsin⁡(2x+α)+sin⁡αdx\int \frac{\sec x}{\sqrt{\sin (2 x+\alpha)+\sin \alpha}} d x is equal to
View answer and explanation
Correct option: A
2sec⁡α(tan⁡x+tan⁡α)+C\sqrt{2 \sec \alpha(\tan x+\tan \alpha)}+C
Option A is correct because after applying trigonometric identities and simplifying, the integral evaluates to the given expression involving tan⁡x\tan x and tan⁡α\tan \alpha.
19 Easy
19 of 746
Analyzing +4 | -1
∫1−7cos⁡2xsin⁡7xcos⁡2xdx=f(x)(sin⁡x)7+C\int \frac{1-7 \cos ^2 x}{\sin ^7 x \cos ^2 x} d x=\frac{f(x)}{(\sin x)^7}+C, then f(x)f(x) is equal to
View answer and explanation
Correct option: C
tan⁡x\tan x
Option C (tan⁡x\tan x) is correct because integrating the given expression results in a function of the form f(x)/(sin x)7 + C, where f(x) = tan x.
20 Easy
20 of 746
Application +4 | -1
The integral ∫etan⁡−1x(1+x+x21+x2)dx\int e^{\tan ^{-1} x}\left(\frac{1+x+x^2}{1+x^2}\right) \mathrm{d} x is equal to
View answer and explanation
Correct option: C
xetan⁡−1x+Cx e^{\tan ^{-1} x}+C
Option C (xetan⁡−1x+Cx e^{\tan ^{-1} x}+C) is correct because the integral can be solved using substitution and the formula ∫ex(f(x)+f′(x))dx=exf(x)+C\int e^x(f(x)+f'(x)) dx = e^x f(x) + C.
21 Easy
21 of 746
Application +4 | -1
∫emtan⁡−1x1+x2dx\int \frac{e^{m \tan ^{-1} x}}{1+x^2} d x equals to
View answer and explanation
Correct option: C
1memtan⁡−1x\frac{1}{m} e^{m \tan ^{-1} x}
Option C (1memtan⁡−1x\frac{{1}}{{m}} e^{{m \tan^{{-1}} x}}) is correct because the integral can be solved by substitution.
22 Easy
22 of 746
Application +4 | -1
The antiderivative of 3x1−9x\frac{{{3^x}}}{{\sqrt {1 - {9^x}} }} with respect to xx is
View answer and explanation
Correct option: A
(log⁡3  e)  sin⁡−1  (3x)+C\left( {{{\log }_3}\,\,e} \right)\,\,{\sin ^{ - 1}}\,\,\left( {{3^x}} \right) + C
Option A ((log⁡3 e) sin⁡−1 (3x)+C\left( {{{\log }_3}\,e} \right)\,{\sin ^{ - 1}}\,\left( {{3^x}} \right) + C) is correct because it is the antiderivative obtained by using substitution and the formula for the integral of 1a2−x2\frac{1}{\sqrt{a^2 - x^2}}.
23 Easy
23 of 746
Application +4 | -1
The value of ∫xexdx\int {x{e^x}dx} is equal to
View answer and explanation
Correct option: B
xex−ex+cx{e^x} - {e^x} + c
Option B (xex−ex+cx{e^x} - {e^x} + c) is correct because it is the result of integrating xexx{e^x} using integration by parts.
24 Easy
24 of 746
Application +4 | -1
If In=∫(log⁡x)ndxI_n=\int(\log x)^n d x, then In+nIn−1=I_n+n I_{n-1}=
View answer and explanation
Correct option: B
x(log⁡x)nx(\log x)^n
Option B (x(log⁡x)nx(\log x)^n) is correct because integration by parts simplifies the expression to the given result.
25 Easy
25 of 746
Application +4 | -1
The integral I=∫e3log⁡e2x+5e2log⁡e2xe4log⁡ex+5e3log⁡ex−7e2log⁡exdxI=\int \frac{e^{3 \log _e 2 x}+5 e^{2 \log _e 2 x}}{e^{4 \log _e x+5 e^{3 \log _e x}-7 e^{2 \log _e x}}} d x is equal to
View answer and explanation
Correct option: B
4ln⁡∣x2+5x−7∣+C4 \ln \left|x^2+5 x-7\right|+C
Option B (4ln⁡∣x2+5x−7∣+C4 \ln \left|x^2+5 x-7\right|+C) is correct because after simplifying the integral, the result matches this option.
26 Easy
26 of 746
Application +4 | -1
∫1sin⁡4x+cos⁡4xdx\int \frac{1}{\sin ^4 x+\cos ^4 x} d x equals
View answer and explanation
Correct option: A
12tan⁡−1(tan⁡2x2)+C\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{\tan 2 x}{\sqrt{2}}\right)+C
Option A is correct because the integral simplifies to a form that can be solved using the arctangent function, resulting in the given expression.
27 Easy
27 of 746
Analyzing +4 | -1
Let f(x)=∫x(1+x)2dx (x≥0)f(x)=\int \frac{\sqrt{x}}{(1+x)^2} d x \, (x \geq 0). Then, f(3)−f(1)f(3)-f(1) is equal to
View answer and explanation
Correct option: C
π12+12−34\frac{\pi}{12}+\frac{1}{2}-\frac{\sqrt{3}}{4}
Option C (π12+12−34\frac{\pi}{12}+\frac{1}{2}-\frac{\sqrt{3}}{4}) is correct because after using the substitution x=tan⁡2θx = \tan^2 \theta and evaluating the definite integral, we get the result in option C.
28 Easy
28 of 746
Application +4 | -1
∫sec⁡2xdxtan⁡2x+4=\int \frac{\sec ^2 x d x}{\sqrt{\tan ^2 x+4}}=
View answer and explanation
Correct option: A
log⁡[tan⁡x+tan⁡2x+4]+c\log \left[\tan x+\sqrt{\tan ^2 x+4}\right]+c
Option A (log⁡[tan⁡x+tan⁡2x+4]+c\log \left[\tan x+\sqrt{\tan ^2 x+4}\right]+c) is correct because it is the result of a u-substitution and applying the standard integral of 1t2+a2\frac{1}{\sqrt{t^2+a^2}}.
29 Easy
29 of 746
Analyzing +4 | -1
If ∫5x+4x2+2x+5dx=(x2+2x+5)1/5+f(x)+c\int \frac{5 x+4}{\sqrt{x^2+2 x+5}} d x=\left(x^2+2 x+5\right)^{1 / 5}+f(x)+c then f(x)f (x) is equal to
View answer and explanation
Correct option: B
−log⁡e(x+1+x2+2x+5)-\log _e\left(x+1+\sqrt{x^2+2 x+5}\right)
The correct answer is B (−log⁡e(x+1+x2+2x+5)-\log _e(x+1+\sqrt{x^2+2 x+5})) because it results from correctly evaluating the integral after completing the square and using a suitable substitution.
30 Easy
30 of 746
Analyzing +4 | -1
∫  dxcos⁡3  2  sin⁡  2x\int \,\,\frac{{dx}}{{{{\cos }^3}\,\,\sqrt {2\,\,\sin \,\,2x} }} is equal to
View answer and explanation
Correct option: A
tan⁡  x   +  tan⁡5/2x5  +  c\sqrt {\tan \,\,x} \,\,\, + \,\,\frac{{{{\tan }^{5/2}}x}}{5}\,\, + \,\,c
Option A (tan⁡,,x,,+,tan⁡5/2x5,+,,c\sqrt {\tan \\,\\,x} \\,\\,\\ + \\,\frac{{{{\tan }^{5/2}}x}}{5}\\,\\ + \\,\\,c) is correct because it is the result of the integration after appropriate trigonometric manipulation and substitution.
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