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JEE • MATHEMATICS • INDEFINITE INTEGRATION

Indefinite integration practice · Mixed difficulty

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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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31 Easy
31 of 995
Application +4 | -1
∫  cos⁡  2x  +  2  sin⁡2  xcos⁡2  x  dx\int \,\,\frac{{\cos \,\,2x\,\, + \,\,2\,\,{{\sin }^2}\,\,x}}{{{{\cos }^2}\,\,x}}\,\,dx is equal to
View answer and explanation
Correct option: C
tan⁡  x  +  c\tan \,\,x\,\, + \,\,c
Option C (tan x + c) is correct because simplifying the integrand using trigonometric identities results in sec²x, whose integral is tan x + c.
32 Easy
32 of 995
Application +4 | -1
∫1x3[log⁡xx]2dx=\int \frac{1}{x^3}\left[\log x^x\right]^2 d x=
View answer and explanation
Correct option: B
13(log⁡x)3+c\frac{1}{3}(\log x)^3+c
Option B (13(log⁡x)3+c\frac{1}{3}(\log x)^3+c) is correct because it is the result of simplifying the expression and then applying a u-substitution.
33 Medium
33 of 995
Analyzing +4 | -1
If ∫cos⁡4xsin⁡2xdx=Acot⁡x+Bsin⁡2x+C2x+D\int \frac{\cos ^4 x}{\sin ^2 x} d x=A \cot x+B \sin 2 x+\frac{C}{2} x+D, then
View answer and explanation
Correct option: B
B=−1/4,C=−3B=-1 / 4, C=-3
Option B is correct because after evaluating the integral, we find that B=−14B = -\frac{1}{4} and C=−3C = -3.
34 Easy
34 of 995
Application +4 | -1
∫dxsin⁡x+cos⁡x\int \frac{\mathrm{d} x}{\sin x+\cos x} is equal to
View answer and explanation
Correct option: C
12log⁡∣tan⁡(x2+π8)∣+C\frac{1}{\sqrt{2}} \log \left|\tan \left(\frac{x}{2}+\frac{\pi}{8}\right)\right|+C
Option C is correct because the integral of 1/(sin x + cos x) can be solved by rewriting the denominator and using trigonometric identities.
35 Easy
35 of 995
Application +4 | -1
∫xx2+1(2ln⁡x+1)dx\int x^{x^2+1}(2 \ln x+1) d x
View answer and explanation
Correct option: D
(xx)x+C\left(x^x\right)^x+C
Option D ((xx)x+C\left(x^x\right)^x+C) is correct because it results from correctly identifying the substitution and integrating the expression.
36 Easy
36 of 995
Understanding +4 | -1
If ∫dx(x2+a2)2\int \frac{d x}{\left(x^2+a^2\right)^2} =1ka2{xx2+a2+1atan⁡−1xa}+C=\frac{1}{k a^2}\left\{\frac{x}{x^2+a^2}+\frac{1}{a} \tan ^{-1} \frac{x}{a}\right\}+C. Then the value of kk, is
View answer and explanation
Correct option: B
2
Option B (2) is correct because by integrating ∫dx(x2+a2)2\int \frac{d x}{\left(x^2+a^2\right)^2} and comparing with the given expression, the value of kk is found to be 2.
37 Medium
37 of 995
Application +4 | -1
∫sin⁡51x(sin⁡x)49dx=\int \sin 51 x(\sin x)^{49} d x=
View answer and explanation
Correct option: A
sin⁡50x(sin⁡x)5050+c\frac{\sin 50 x(\sin x)^{50}}{50}+c
Option A (sin⁡50x(sin⁡x)5050+c\frac{{\sin 50 x(\sin x)^{{50}}}}{{50}}+c) is correct because the integral simplifies to this form after applying trigonometric identities and integration by parts.
38 Easy
38 of 995
Analyzing +4 | -1
∫dxx(x5+1)\int \frac{{dx}}{{x\left( {{x^5} + 1} \right)}} is equal to
View answer and explanation
Correct option: C
15  log⁡  (x5x5+  1) +c\frac{1}{5}\,\,\log \,\,\left( {\frac{{{x^5}}}{{{x^5} + \,\,1}}} \right)\, + c
Option C is correct because the integral can be solved using substitution and partial fractions.
39 Easy
39 of 995
Application +4 | -1
∫dxx+xlog⁡x=\int \frac{d x}{x+x \log x}=
View answer and explanation
Correct option: A
log⁡(1+log⁡x)\log (1+\log x)
Option A is correct because the integral can be rewritten and solved using a substitution, resulting in log⁡(1+log⁡x)\log(1+\log x).
40 Easy
40 of 995
Application +4 | -1
∫e2x+log⁡xdx=\int e^{2 x+\log x} d x=
View answer and explanation
Correct option: A
14(2x−1)e2x+c\frac{1}{4}(2 x-1) e^{2 x}+c
Option A (14(2x−1)e2x+c\frac{1}{4}(2 x-1) e^{2 x}+c) is correct because it is the result of integrating e2x+log⁡xe^{2x+\log x} using integration by parts.
41 Easy
41 of 995
Application +4 | -1
Let f(x)f(x) be a polynomial satisfying f(0)=2,f′(0)=3f(0)=2, f^{\prime}(0)=3 and f′′(x)=f(x)f^{\prime \prime}(x)=f(x). Then f(4)f(4) is equal to
View answer and explanation
Correct option: B
5(e8−1)2e4\frac{5\left(e^8-1\right)}{2 e^4}
Option A is incorrect. Option B is correct because by solving the differential equation and using the given initial conditions, we find f(4)=5(e8−1)2e4f(4)=\frac{5(e^8-1)}{2 e^4}.
42 Easy
42 of 995
Application +4 | -1
∫  (x+1)2x(x2+1)  dx\int \,\,\frac{{{{\left( {x + 1} \right)}^2}}}{{x\left( {{x^2} + 1} \right)}}\,\,dx is equal to
View answer and explanation
Correct option: B
log⁡e  x  +  2  tan⁡−1  x  +  c{\log _e}\,\,x\,\, + \,\,2\,\,{\tan ^{ - 1}}\,\,x\,\, + \,\,c
Option B (log⁡e  x  +  2  tan⁡−1  x  +  c{\log _e}\,\,x\,\, + \,\,2\,\,{\tan ^{ - 1}}\,\,x\,\, + \,\,c) is correct because the integral simplifies to the sum of the integral of 1/x and 2/(x2+1).
43 Medium
43 of 995
Analyzing +4 | -1
The value of ∫cos⁡xsin⁡2x(sin⁡x+cos⁡x)dx\int \frac{\cos x}{\sin ^2 x(\sin x+\cos x)} d x is equal to
View answer and explanation
Correct option: A
log⁡∣1+tan⁡xtan⁡x∣−cot⁡x+C\log \left|\frac{1+\tan x}{\tan x}\right|-\cot x+C
Option A (log⁡∣1+tan⁡xtan⁡x∣−cot⁡x+C\log \left|\frac{{1+\tan x}}{{\tan x}}\right|-\cot x+C) is correct because the integral can be solved by dividing the numerator and denominator by cos3(x) and using partial fractions after substituting tan(x) = t.
44 Medium
44 of 995
Application +4 | -1
∫e−2xsin⁡3xdx=\int e^{-2 x} \sin 3 x d x=
View answer and explanation
Correct option: D
−113e−2x[2sin⁡3x+3cos⁡3x]+c-\frac{1}{13} e^{-2 x}[2 \sin 3 x+3 \cos 3 x]+c
Option D (−113e−2x[2sin⁡3x+3cos⁡3x]+c-\frac{{1}}{{13}} e^{-2 x}[2 \sin 3 x+3 \cos 3 x]+c) is correct because it is the result of integrating the given expression using integration by parts twice.
45 Easy
45 of 995
Application +4 | -1
If ∫dx5−4sin⁡x+3cos⁡x\int \frac{d x}{5-4 \sin x+3 \cos x} is equal to 12+f(x)+C\frac{1}{2+f(x)}+C then f(x)f(x) is equal to
View answer and explanation
Correct option: C
−tan⁡x2-\tan \frac{x}{2}
Option C is correct because after applying the tangent half-angle substitution and simplifying, the integral evaluates to 12−tan⁡x2+C\frac{1}{2-\tan \frac{x}{2}}+C, thus f(x)=−tan⁡x2f(x) = -\tan \frac{x}{2}.

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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
46 Easy
46 of 995
Analyzing +4 | -1
∫e−x(1−tan⁡x)sec⁡x dx\int e^{-x}(1-\tan x) \sec x\, d x is equal to
View answer and explanation
Correct option: D
None of these
Option D is correct because after applying the integration rule ∫ekx[kf(x)+f′(x)]dx=ekxf(x)\int e^{kx} [kf(x) + f'(x)] dx = e^{kx}f(x), we get −e−xsec⁡x+C-e^{-x} \sec x + C.
47 Easy
47 of 995
Application +4 | -1
∫x−2x(2log⁡x−x)dx=\int \frac{x-2}{x(2 \log x-x)} d x=
View answer and explanation
Correct option: B
log⁡(12log⁡x−x)+c\log \left(\frac{1}{2 \log x-x}\right)+c
Option B is correct because the integral can be solved by rearranging the integrand and using substitution.
48 Easy
48 of 995
Application +4 | -1
∫dxe−2x(e2x+1)2=\int \frac{d x}{e^{-2 x}\left(e^{2 x}+1\right)^2}=
View answer and explanation
Correct option: A
−12(e2x+1)+c\frac{-1}{2\left(e^{2 x}+1\right)}+c
Option A (−12(e2x+1)+c\frac{-1}{2\left(e^{2 x}+1\right)}+c) is correct because the integral simplifies to this form after substitution.
49 Easy
49 of 995
Analyzing +4 | -1
The value of ∫(1+x2)dx(1−x2)1+x2+x4\int \frac{\left(1+x^2\right) d x}{\left(1-x^2\right) \sqrt{1+x^2+x^4}}, is
View answer and explanation
Correct option: A
−123log⁡∣x4+x2+1−3xx4+x2+1+3x∣+C-\frac{1}{2 \sqrt{3}} \log \left|\frac{\sqrt{x^4+x^2+1}-\sqrt{3} x}{\sqrt{x^4+x^2+1}+\sqrt{3} x}\right|+C
Option A is the correct answer because the integral evaluates to −123log⁡∣x4+x2+1−3xx4+x2+1+3x∣+C-\frac{1}{2 \sqrt{3}} \log \left|\frac{\sqrt{x^4+x^2+1}-\sqrt{3} x}{\sqrt{x^4+x^2+1}+\sqrt{3} x}\right|+C after appropriate substitutions and simplification.
50 Easy
50 of 995
Application +4 | -1
The value of ∫log⁡e(x+x2+1)x2+1dx\int \frac{\log _e\left(x+\sqrt{x^2+1}\right)}{\sqrt{x^2+1}} d x, is
View answer and explanation
Correct option: B
{log⁡e(x+x2+1)}2+C\left\{\log _e\left(x+\sqrt{x^2+1}\right)\right\}^2+C
Option B is correct because the integral is solved by recognizing the derivative of the logarithmic term and applying a simple substitution.
51 Medium
51 of 995
Application +4 | -1
If ∫dxx(1+x4)10=a(1+x4)−8+b(1+x4)−9\int \frac{d x}{\sqrt{x}(1+\sqrt[4]{x})^{10}}=a(1+\sqrt[4]{x})^{-8}+b(1+\sqrt[4]{x})^{-9} then a+ba+b is
View answer and explanation
Correct option: B
−118-\frac{1}{18}
Option B (−118-\frac{1}{18}) is correct because after evaluating the integral and comparing coefficients, a+ba+b equals this value.
52 Medium
52 of 995
Application +4 | -1
∫1  +  xx  +  e−x    dx\int \frac{{1\,\, + \,\,x}}{{x\,\, + \,\,{e^{ - x}}}}\,\,\;dx is equal to
View answer and explanation
Correct option: C
log⁡  ∣ (1  +  xex) ∣+c\log \,\,\left| {\,\left( {1\,\, + \,\,x{e^x}} \right)\,} \right| + c
Option C (log⁡ ∣ (1+xex) ∣+c\log \,|\,(1 + x{e^x})\,| + c) is correct because by using substitution, the integral simplifies to this logarithmic form.
53 Hard
53 of 995
Analyzing +4 | -1
∫x4−  1x2(x4+x2+1)1/2  dx\int \frac{{{x^4} - \,\,1}}{{{x^2}{{\left( {{x^4} + {x^2} + 1} \right)}^{1/2}}}}\;dx is equal to
View answer and explanation
Correct option: D
x4+x2+1x+c\frac{{\sqrt {{x^4} + {x^2} + 1} }}{x} + c
The correct answer is D (x4+x2+1x+c\frac{{\sqrt {{{{x^4} + {x^2} + 1}}} }}{x} + c) because the integral simplifies to the derivative of the given expression.
54 Medium
54 of 995
Application +4 | -1
∫sin⁡2xcos⁡3xdx=\int \sin 2 x \cos 3 x d x=
View answer and explanation
Correct option: B
12(cos⁡x−15cos⁡5x)+c\frac{1}{2}\left(\cos x-\frac{1}{5} \cos 5 x\right)+c
Option B is correct because the integral of sin⁡2xcos⁡3x\sin 2x \cos 3x is solved using the product-to-sum trigonometric identities.
55 Easy
55 of 995
Application +4 | -1
∫sin⁡2xsin⁡5xsin⁡3xdx=\int \frac{\sin 2 x}{\sin 5 x \sin 3 x} d x=
View answer and explanation
Correct option: C
13log⁡sin⁡3x−15log⁡sin⁡5x+c\frac{1}{3} \log \sin 3 x-\frac{1}{5} \log \sin 5 x+c
Option A (log⁡sin⁡3x−log⁡sin⁡5x+c\log \sin 3 x-\log \sin 5 x+c) is incorrect. Option A is wrong because the coefficients 1/3 and 1/5 are missing.
56 Easy
56 of 995
Application +4 | -1
Let f(x)f(x) be a polynomial satisfying f(0)=2f(0)=2, f′(0)=3f^{\prime}(0)=3 and f′′(x)=f(x)f^{\prime \prime}(x)=f(x). Then, f(4)=f(4)=
View answer and explanation
Correct option: B
5(e8−1)2e4\frac{5\left(e^8-1\right)}{2 e^4}
Option B is correct. The polynomial f(x)f(x) is found by solving the differential equation f′′(x)=f(x)f^{\prime \prime}(x) = f(x) with given initial conditions, and then evaluating f(4)f(4).
57 Easy
57 of 995
Application +4 | -1
∫x+(cos⁡−13x)21−9x2dx\int \frac{x+\left(\cos ^{-1} 3 x\right)^2}{\sqrt{1-9 x^2}} d x is equal to
View answer and explanation
Correct option: A
C−(191−9x2+(cos⁡−13x)3)C-\left(\frac{1}{9} \sqrt{1-9 x^2}+\left(\cos ^{-1} 3 x\right)^3\right)
Option A (\C−(191−9x2+(cos⁡−13x)3)C - (\frac{1}{9} \sqrt{1-9x^2} + (\cos^{-1} 3x)^3)\\) is correct. The integral is solved by splitting it into two parts and using substitution for each part.
58 Medium
58 of 995
Application +4 | -1
∫xx4−1dx=\int \frac{x}{x^4-1} d x=
View answer and explanation
Correct option: A
14log⁡[x2−1x2+1]+c\frac{1}{4} \log \left[\frac{x^2-1}{x^2+1}\right]+c
Option A is correct because the integral is solved by substituting t=x2t = x^2 and using partial fraction decomposition.
59 Medium
59 of 995
Application +4 | -1
3a  ∫01  (ax−1a−1)2dx3a\,\,\int \nolimits_0^1 \,\,{\left( {\frac{{ax - 1}}{{a - 1}}} \right)^2}dx is equal to
View answer and explanation
Correct option: A
a−1+(a−1)−2a - 1 + {\left( {a - 1} \right)^{ - 2}}
Option A is correct because the integral evaluates to the given expression after simplification, matching the form in option A.
60 Easy
60 of 995
Application +4 | -1
Let P(x)P(x) be a polynomial of degree nn with leading coefficient 1. Let v(x)v(x) be any function and v1(x)=v_1(x)= ∫v(x)dx\int v(x) \mathrm{d} x
v2(x)=∫v1(x)dx…vn+1(x)=∫vn(x)dx.v_2(x)=\int v_1(x) \mathrm{d} x \ldots v_{n+1}(x)=\int v_n(x) \mathrm{d} x .
then ∫P(x)v(x)dx\int P(x) v(x) \mathrm{d} x is equal to
View answer and explanation
Correct option: B
P(x)v1(x)−P′(x)v2(x)+P′′(x)v3(x)…+(−1)nP(x) v_1(x)-P^{\prime}(x) v_2(x)+P^{\prime \prime}(x) v_3(x) \ldots+(-1)^n n!vn+1(x)n ! v_{n+1}(x)
The correct answer is B because it represents the general formula for integration by parts applied repeatedly to a polynomial and a function.
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