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JEE • MATHEMATICS • INDEFINITE INTEGRATION

Indefinite integration practice · Easy

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Practice format and chapter quiz

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 541-570 of 746 questions

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541 Easy
541 of 746
Application +4 | -1
∫14cos⁡3x−3cos⁡xdx\int \frac{1}{4 \cos ^3 x-3 \cos x} d x is equal to
View answer and explanation
Correct option: B
13ln⁡∣sec⁡3x+tan⁡3x∣+c\frac{1}{3} \ln |\sec 3 x+\tan 3 x|+c
Option B is correct because the integral simplifies to 13ln⁡∣sec⁡3x+tan⁡3x∣+c\frac{1}{3} \ln |\sec 3 x+\tan 3 x|+c after using the triple angle formula and integrating.
542 Easy
542 of 746
Application +4 | -1
∫  cos⁡  [2  cot⁡−1  1  −  x1  +  x]  dx\int \,\,\cos \,\,\left[ {2\,\,{{\cot }^{ - 1}}\,\,\sqrt {\frac{{1\,\, - \,\,x}}{{1\,\, + \,\,x}}} } \right]\,\,dx is equal to
View answer and explanation
Correct option: C
−  12  x2  +  c- \,\,\frac{1}{2}\,\,{x^2}\,\, + \,\,c
Option C (−12x2+c- \frac{1}{2}\\ {x^2}\\ + \\ c) is correct because the expression inside the cosine simplifies to -x, and the integral of -x is -x2/2 + c.
543 Easy
543 of 746
Analyzing +4 | -1
∫02  2x  −  22x  −  x2  dx\int \nolimits_0^2 \,\,\frac{{2x\,\, - \,\,2}}{{2x\,\, - \,\,{x^2}}}\,\,dx is equal to
View answer and explanation
Correct option: A
0
Option A (0) is correct because substituting t=2x−x2t = 2x - x^2 makes the limits of integration the same, resulting in a zero integral.
544 Easy
544 of 746
Application +4 | -1
∫1+tan⁡xx+log⁡sec⁡xdx=\int \frac{1+\tan x}{x+\log \sec x} d x=
View answer and explanation
Correct option: A
log⁡(x+log⁡sec⁡x)+c\log (x+\log \sec x)+c
Option A is correct because the integral simplifies to log⁡(x+log⁡sec⁡x)+c\log(x + \log \sec x) + c by recognizing the numerator as the derivative of the denominator.
545 Easy
545 of 746
Application +4 | -1
If f(x)=log⁡xx3f(x)=\frac{\log x}{x^3}, then its antiderivative F(x)F(x) given by
View answer and explanation
Correct option: A
C−12x2log⁡(xe)C-\frac{1}{2 x^2} \log (x \sqrt{e})
Option A (C−12x2log⁡(xe)C-\frac{1}{2 x^2} \log (x \sqrt{e})) is correct because it is the result of integrating the given function using integration by parts.
546 Easy
546 of 746
Understanding +4 | -1
If nn is an odd positive integer, then ∫∣xn∣dx\int\left|x^n\right| d x is equal to
View answer and explanation
Correct option: C
∣xn∣xn+1+C\frac{\left|x^n\right| x}{n+1}+C
Option C is correct because, for odd n, the integral of |x^n| dx results in an expression that involves |x|^n * x / (n+1) + C, accounting for both positive and negative values of x.
547 Easy
547 of 746
Application +4 | -1
∫sec⁡2/3xcosec⁡4/3xdx=\int \sec ^{2 / 3} x \operatorname{cosec}^{4 / 3} x d x=
View answer and explanation
Correct option: B
−3(tan⁡x)−1/3+c-3(\tan x)^{-1 / 3}+c
Option B is correct because the integral can be simplified using trigonometric identities and solved using substitution, resulting in −3(tan⁡x)−1/3+c-3(\tan x)^{-1 / 3}+c.
548 Easy
548 of 746
Application +4 | -1
∫11+3sin⁡2xdx\int \frac{1}{1+3 \sin ^2 x} d x is equal to
View answer and explanation
Correct option: B
12tan⁡−1(2tan⁡x)+C\frac{1}{2} \tan ^{-1}(2 \tan x)+C
Option B, 12tan⁡−1(2tan⁡x)+C\frac{1}{2} \tan ^{-1}(2 \tan x)+C, is correct because dividing by cos⁡2x\cos^2 x and using a u-substitution leads to this result.
549 Easy
549 of 746
Analyzing +4 | -1
∫x−3{sin⁡−1(ln⁡x)+cos⁡−1(ln⁡x)}dx\int \sqrt{x-3}\left\{\sin ^{-1}(\ln x)+\cos ^{-1}(\ln x)\right\} d x is equal to
View answer and explanation
Correct option: C
Does not exist
The correct answer is D (Does not exist) because the domain of sin⁡−1(ln⁡x)+cos⁡−1(ln⁡x)\sin^{-1}(\ln x) + \cos^{-1}(\ln x) is restricted to −1≤ln⁡x≤1-1 \leq \ln x \leq 1, which is not always true.
550 Easy
550 of 746
Application +4 | -1
∫  1  +  x  +  x  +  x2x  +  1  +  x  dx\int \,\,\frac{{1\,\, + \,\,x\,\, + \,\,\sqrt {x\,\, + \,\,{x^2}} }}{{\sqrt x \,\, + \,\,\sqrt {1\,\, + \,\,x} }}\,\,dx is equal to
View answer and explanation
Correct option: B
23  (1+x)3/2+  c\frac{2}{3}\,\,{\left( {1 + x} \right)^{3/2}} + \,\,c
Option B (23 (1+x)3/2+ c\frac{2}{3}\,{\left( {1 + x} \right)^{3/2}} + \,c) is correct because the integral simplifies to ∫1+xdx\int \sqrt{1+x} dx, which integrates to the given expression.
551 Easy
551 of 746
Application +4 | -1
∫  1  +  tan⁡  xe−x  cos⁡  x  dx\int \,\,\frac{{1\,\, + \,\,\tan \,\,x}}{{{e^{ - x}}\,\,\cos \,\,x}}\,\,dx is equal to
View answer and explanation
Correct option: C
ex  sec⁡  x+  c{e^x}\,\,\sec \,\,x + \,\,c
Option C (exsec⁡,x+,c{e^x}\sec \\,x + \\,c) is correct because the integral simplifies to this expression after applying integration by parts.
552 Easy
552 of 746
Application +4 | -1
If ∫f(x)dx=F(x)\int f(x) d x=F(x), then ∫x3f(x2)dx\int x^3 f\left(x^2\right) d x is equal to
View answer and explanation
Correct option: B
12[x2F(x2)−∫F(x2)d(x2)]\frac{1}{2}\left[x^2 F\left(x^2\right)-\int F\left(x^2\right) d\left(x^2\right)\right]
Option B is correct. The integral is solved using substitution and integration by parts, resulting in the given expression.
553 Easy
553 of 746
Application +4 | -1
If ∫(e2x+2ex−e−x−1) e(ex+e−x)dx=g(x)e(ex+e−x)+c,\int {\left( {{e^{2x}} + 2{e^x} - {e^{ - x}} - 1} \right)} \,{e^{({e^x} + {e^{ - x}})}}dx = g\left( x \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}} + c, where cc is a constant of integration, then g(0)g\left( 0 \right) is equal to
View answer and explanation
Correct option: A
2
Option A (2) is correct because after solving the integral and substituting x=0 into the resulting function g(x), we obtain the value 2.
554 Easy
554 of 746
Application +4 | -1
∣ln⁡x∣dx|\ln x| \mathrm{d} x equals (0<x<1)(0 < x < 1)
View answer and explanation
Correct option: A
x+x∣ln⁡x∣+cx+x|\ln x|+c
Option B (x∣ln⁡x∣−x+cx|\ln x|-x+c) is correct because integrating |ln x| with the given condition 0 < x < 1 results in the expression x|ln x| - x + c.
555 Easy
555 of 746
Application +4 | -1
∫tan⁡−11−x1+xdx=\int \tan ^{-1} \sqrt{\frac{1-x}{1+x}} d x=
View answer and explanation
Correct option: A
12(xcos⁡−1x−1−x2)+C\frac{1}{2}\left(x \cos ^{-1} x-\sqrt{1-x^2}\right)+C
Option A (12(xcos⁡−1x−1−x2)+C\frac{1}{2}\left(x \cos ^{-1} x-\sqrt{1-x^2}\right)+C) is correct because it is the result of integrating the given expression using trigonometric substitution and integration by parts.
556 Easy
556 of 746
Application +4 | -1
∫ex(x+2x+4)2dx=\int e^x\left(\frac{x+2}{x+4}\right)^2 d x=
View answer and explanation
Correct option: A
xexx+4+c\frac{x e^x}{x+4}+c
Option A (xexx+4+c\frac{x e^x}{x+4}+c) is correct because the integral is solved using the form ∫ex[f(x)+f′(x)]dx=exf(x)+c\int e^x [f(x) + f'(x)] dx = e^x f(x) + c.
557 Easy
557 of 746
Application +4 | -1
If b>ab>a, and I=∫abx−ab−xdxI=\int_a^b \sqrt{\frac{x-a}{b-x}} d x, then I equals
View answer and explanation
Correct option: A
π2(b−a)\frac{\pi}{2}(b-a)
Option A (π2(b−a)\frac{\pi}{2}(b-a)) is correct because the integral evaluates to this value after a suitable substitution and trigonometric simplification.
558 Easy
558 of 746
Application +4 | -1
∫cos⁡5x+cos⁡4x1−2cos⁡3xdx\int \frac{\cos 5 x+\cos 4 x}{1-2 \cos 3 x} d x is equal to
View answer and explanation
Correct option: C
−sin⁡2x2−sin⁡x+c-\frac{\sin 2 x}{2}-\sin x+c
Option C (-sin⁡2x2−sin⁡x+c\frac{{\sin 2x}}{{2}}-\sin x+c) is correct because it is the simplified result after applying trigonometric identities and integration.
559 Easy
559 of 746
Application +4 | -1
∫(x2010+x804+x402)(2x1608+5x402+10)1402dx=110a(2x2010+5x804+10x402)a402, then a−400 is \begin{aligned} & \int\left(x^{2010}+x^{804}+x^{402}\right)\left(2 x^{1608}+5 x^{402}+10\right)^{\frac{1}{402}} d x \\ & =\frac{1}{10 a}\left(2 x^{2010}+5 x^{804}+10 x^{402}\right)^{\frac{a}{402}}, \text { then } a-400 \text { is } \end{aligned}
View answer and explanation
Correct option: A
3
Option A (3) is correct because it is the value of a - 400, derived from the given integral by substitution and simplification.
560 Easy
560 of 746
Application +4 | -1
∫(2+sin⁡2x1+cos⁡2x)exdx=\int\left(\frac{2+\sin 2 x}{1+\cos 2 x}\right) e^x d x=
View answer and explanation
Correct option: D
extan⁡x+ce^x \tan x+c
Option D (extan⁡x+ce^x \tan x+c) is correct because the integral simplifies to this expression by recognizing the form of ∫ex[f(x)+f′(x)]dx\int e^x [f(x) + f'(x)] dx.
561 Easy
561 of 746
Application +4 | -1
∫sin⁡xcos⁡xacos⁡2x+bsin⁡2xdx=\int \frac{\sin x \cos x}{a \cos ^2 x+b \sin ^2 x} d x=
View answer and explanation
Correct option: A
12(b−a)log⁡(acos⁡2x+bsin⁡2x)+c\frac{1}{2(b-a)} \log \left(a \cos ^2 x+b \sin ^2 x\right)+c
Option A is correct because the integral can be solved using substitution.
562 Easy
562 of 746
Application +4 | -1
If ∫exsin⁡xdx=12ex⋅a+c\int e^x \sin x d x=\frac{1}{2} e^x \cdot a+c, then a=a=
View answer and explanation
Correct option: A
sin⁡x−cos⁡x\sin x-\cos x
Option A (sin⁡x−cos⁡x\sin x-\cos x) is correct because it is the result of solving the integral ∫exsin⁡xdx\int e^x \sin x dx using integration by parts twice and comparing with the given form.
563 Easy
563 of 746
Application +4 | -1
∫x3log⁡xdx=\int x^3 \log x d x=
View answer and explanation
Correct option: B
116[4x4log⁡x−x4]+c\frac{1}{16}\left[4 x^4 \log x-x^4\right]+c
Option B is correct because the integral of x3 log(x) dx is found using integration by parts, resulting in 1/16[4x4 log(x) - x4] + c.
564 Easy
564 of 746
Application +4 | -1
∫sin⁡2xlog⁡ecos⁡xdx\int \sin 2 x \log _e \cos x d x is equal to
View answer and explanation
Correct option: C
(12−log⁡ecos⁡x)cos⁡2x+C\left(\frac{1}{2}-\log _e \cos x\right) \cos ^2 x+C
Option C, (12−log⁡ecos⁡x)cos⁡2x+C\left(\frac{1}{2}-\log _e \cos x\right) \cos ^2 x+C, is correct because it results from integrating sin⁡2xlog⁡ecos⁡x\sin 2x \log_e \cos x using integration by parts and substitution.
565 Easy
565 of 746
Application +4 | -1
∫  x  −  1x  +  1  dx\int \,\,\sqrt {\frac{{x\,\, - \,\,1}}{{x\,\, + \,\,1}}\,\,dx} is equal to
View answer and explanation
Correct option: B
x2−  1−  sin⁡−1  x+c\sqrt {{x^2} - \,\,1} - \,\,{\sin ^{ - 1}}\,\,x + c
Option B (x2− 1− sin⁡−1 x+c\sqrt {{x^2} - \,1} - \,{\sin ^{ - 1}}\,x + c) is correct because it is the result of simplifying the integrand and using appropriate integration techniques.
566 Easy
566 of 746
Application +4 | -1
∫dx(x+1)  4x+3\int \frac{{dx}}{{\left( {x + 1} \right)\,\,\sqrt {4x + 3} }} is equal to
View answer and explanation
Correct option: C
2 tan⁡−1  4x+3  +  c2\,{\tan ^{ - 1}}\,\,\sqrt {4x + 3} \,\, + \,\,c
Option C (2 tan⁡−1 4x + 3 + c2\,{\tan ^{ - 1}}\,\sqrt {4x\, + \,3} \, + \,c) is correct because it is the result of the integration after substitution and simplification.
567 Easy
567 of 746
Understanding +4 | -1
∫  cos⁡  x  +  x  sin⁡  xx2  +  x  cos⁡  x    dx\int \,\,\frac{{\cos \,\,x\,\, + \,\,x\,\,\sin \,\,x}}{{{x^2}\,\, + \,\,x\,\,\cos \,\,x}}\,\,\;dx is equal to
View answer and explanation
Correct option: D
log⁡  ∣ xx  +  cos⁡  x ∣+c\log \,\,\left| {\,\frac{x}{{x\,\, + \,\,\cos \,\,x}}\,} \right| + c
Option D, \log \,\left| {\,\frac{x}{{x\,+\cos \,x}}\} \right| + c\log \,\left| {\,\frac{x}{{x\,+\cos \,x}}\} \right| + c, is correct because the integral simplifies to this form after algebraic manipulation and recognizing the derivative in the denominator.
568 Easy
568 of 746
Application +4 | -1
The value of ∫1+sin⁡x1−sin⁡xdx\int \frac{1+\sin x}{1-\sin x} d x is
View answer and explanation
Correct option: C
2tan⁡(x2+π4)−x+C2 \tan \left(\frac{x}{2}+\frac{\pi}{4}\right)-x+C
Option C (2tan⁡(x2+π4)−x+C2 \tan \left(\frac{x}{2}+\frac{\pi}{4}\right)-x+C) is correct because it is the result of integrating the given expression and simplifying using trigonometric identities.
569 Easy
569 of 746
Analyzing +4 | -1
∫cos⁡x−sin⁡xcos⁡x+sin⁡x(2+2sin⁡2x)dx\int \frac{\cos x-\sin x}{\cos x+\sin x}(2+2 \sin 2 x) d x is equal to
View answer and explanation
Correct option: A
sin⁡2x+C\sin 2 x+C
Option A (sin⁡2x+C\sin 2 x+C) is correct because the integral simplifies to 2∫cos⁡2xdx2 \int \cos 2x dx, which equals sin⁡2x+C\sin 2x + C.
570 Easy
570 of 746
Application +4 | -1
∫extan⁡2(ex)dx=\int e^x \tan ^2\left(e^x\right) d x=
View answer and explanation
Correct option: D
tan⁡(ex)−ex+c\tan \left(e^x\right)-e^x+c
Option D (tan⁡(ex)−ex+c\tan \left(e^x\right)-e^x+c) is correct because the integral simplifies to this form after substitution and integration.
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