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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 31-60 of 746 questions
31 Easy
31 of 746
Analyzing +4 | -1
∫(sin⁡θcos⁡3θ+sin⁡3θcos⁡9θ+sin⁡9θcos⁡27θ)dθ\int\left(\frac{\sin \theta}{\cos 3 \theta}+\frac{\sin 3 \theta}{\cos 9 \theta}+\frac{\sin 9 \theta}{\cos 27 \theta}\right) d \theta is equal to
View answer and explanation
Correct option: C
12log⁡∣sec⁡27θ27sec⁡θ∣+C\frac{1}{2} \log \left|\frac{\sqrt[27]{\sec 27 \theta}}{\sec \theta}\right|+C
Option C (12log⁡∣sec⁡27θ27sec⁡θ∣+C\frac{1}{2} \log \left|\frac{\sqrt[27]{\sec 27 \theta}}{\sec \theta}\right|+C) is correct because after simplifying the trigonometric terms and integrating, the result matches this option.
32 Easy
32 of 746
Application +4 | -1
If ∫x5e−4x3dx=148e−4x3f(x)+C\int {{x^5}{e^{ - 4{x^3}}}dx = \frac{1}{{48}}{e^{ - 4{x^3}}}f\left( x \right) + C}, where CC is a constant of integration, then f(x)f(x) is equal to
View answer and explanation
Correct option: A
−4x3−1- 4{x^3} - 1
Option A (−4x3−1- 4{x^3} - 1) is correct because it is the function f(x)f(x) obtained after performing integration by parts and comparing with the given equation.
33 Easy
33 of 746
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.
34 Easy
34 of 746
Application +4 | -1
Let k(x)=∫(x2+1)dxx3+3x+63\mathrm{k}(\mathrm{x})=\int \frac{\left(x^2+1\right) d x}{\sqrt[3]{x^3+3 x+6}} and k(−1)=123\mathrm{k}(-1)=\frac{1}{\sqrt[3]{2}} then the value of k(−2)\mathrm{k}(-2) is
View answer and explanation
Correct option: A
2
Option A (2) is correct because after evaluating the integral and applying the given condition k(-1), we find the value of the constant and then evaluate k(-2).
35 Easy
35 of 746
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.
36 Easy
36 of 746
Application +4 | -1
The value of ∫sin⁡8x−cos⁡8x1−2sin⁡2xcos⁡2xdx\int \frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x} d x is:
View answer and explanation
Correct option: B
−12sin⁡2x+C-\frac{1}{2} \sin 2 x+C
Option B (−12sin⁡2x+C-\frac{1}{2} \sin 2 x+C) is correct because simplification using trigonometric identities leads to a straightforward integration.
37 Easy
37 of 746
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.
38 Easy
38 of 746
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.
39 Easy
39 of 746
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.
40 Easy
40 of 746
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.
41 Easy
41 of 746
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).
42 Easy
42 of 746
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.
43 Easy
43 of 746
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}.
44 Easy
44 of 746
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).
45 Easy
45 of 746
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 746
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 746
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 746
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 746
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 746
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 Easy
51 of 746
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.
52 Easy
52 of 746
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).
53 Easy
53 of 746
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.
54 Easy
54 of 746
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.
55 Easy
55 of 746
Application +4 | -1
∫xexdx=\int \sqrt{x} e^{\sqrt{x}} d x=
View answer and explanation
Correct option: B
(2x−4x+4)ex+c(2 x-4 \sqrt{x}+4) e^{\sqrt{x}}+c
Option B ((2x−4x+4)ex+c(2 x-4 \sqrt{x}+4) e^{\sqrt{x}}+c) is correct because it is the result of integrating xex\sqrt{x} e^{\sqrt{x}} using substitution and integration by parts twice.
56 Easy
56 of 746
Application +4 | -1
Let f(x)=1+3xlog⁡3f(x)=1+3^x \log 3 and F(x)F(x) be its antiderivative which assume the value 7 for x=2x=2. The value of xx for which the curve F(x)F(x) cuts the abscissa axis is
View answer and explanation
Correct option: B
1
Option B (1) is correct because it is the value of x that makes F(x) = 0, after finding the constant of integration.
57 Easy
57 of 746
Application +4 | -1
The solution of the equation
∫log⁡  2x  1ex−1  dx=log⁡  32\int \limits_{\log \,\,2}^x \,\,\frac{1}{{{e^x} - 1}}\,\,dx = \log \,\,\frac{3}{2}
is given by x=x =
View answer and explanation
Correct option: C
log⁡  4\log \,\,4
Option C (log⁡ 4\log \,4) is correct because solving the definite integral and equating it to log⁡ 32\log \,\frac{3}{2} leads to x=log⁡4x = \log 4.
58 Easy
58 of 746
Application +4 | -1
Let α∈(0,  π/2)\alpha \in \left( {0,\,\,\pi /2} \right) be fixed. If the integral ∫tan⁡x+tan⁡αtan⁡x−tan⁡αdx=A(x)cos⁡2α+B(x)\int {\frac{{\tan x + \tan \alpha }}{{\tan x - \tan \alpha }}dx = A\left( x \right)\cos 2\alpha + B\left( x \right)} sin⁡2α+C \sin 2\alpha + C\,, where CC is a constant of integration, then the functions A(x)A\left( x \right) and B(x)B\left( x \right) are respectively
View answer and explanation
Correct option: B
x−αx - \alpha and log⁡e∣sin⁡(x−α)∣{\log _e}\left| {\sin \left( {x - \alpha } \right)} \right|
Option B (x−αx - \alpha and log⁡e∣sin⁡(x−α)∣{\log _e}\left| {\sin \left( {x - \alpha } \right)} \right|) is correct because these are the functions A(x) and B(x) obtained after evaluating the given integral.
59 Easy
59 of 746
Application +4 | -1
∫xsin⁡2xdx=\int x \sin ^2 x d x=
View answer and explanation
Correct option: D
x24−x4sin⁡2x−18cos⁡2x+c\frac{x^2}{4}-\frac{x}{4} \sin 2 x-\frac{1}{8} \cos 2 x+c
Option D (x24−x4sin⁡2x−18cos⁡2x+c\frac{x^2}{4}-\frac{x}{4} \sin 2 x-\frac{1}{8} \cos 2 x+c) is correct because it is the result of integrating xsin⁡2xx \sin^2 x using trigonometric identities and integration by parts.
60 Easy
60 of 746
Application +4 | -1
A value of α\alpha such that ∫αα+1dx(x+a)(x+α+1)=log⁡e(98)\int\limits_\alpha ^{\alpha + 1} {\frac{{dx}}{{\left( {x + a} \right)\left( {x + \alpha + 1} \right)}} = {{\log }_e}\left( {\frac{9}{8}} \right)} is
View answer and explanation
Correct option: A
−2- 2
Option A (-2) is correct because it is a value of α\alpha that satisfies the given integral equation after solving the integral and substituting the limits.
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