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JEE • MATHEMATICS • INDEFINITE INTEGRATION

Indefinite integration practice · Medium

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31 Medium
31 of 411
Application +4 | -1
Let ddx(F(x))=esin⁡xx,x>0\frac{d}{d x}(F(x))=\frac{e^{\sin x}}{x}, {x} > 0. If ∫142esin⁡x2xdx=F(K)−F(1)\int\limits_{1}^4 \frac{2 e^{\sin x^2}}{x} d x=F(K)-F(1), then the possible value of K{K} is
View answer and explanation
Correct option: C
16
Option C (16) is correct because it is the value of K obtained by comparing the given integral with the definition of F(x).
32 Medium
32 of 411
Application +4 | -1
∫  x3  1  +  x43 7dx\int \,\,\sqrt[3]{x}\,\,\sqrt[7]{{1\,\, + \,\,\sqrt[3]{{{x^4}}}\,}}dx is equal to
View answer and explanation
Correct option: A
2132  {1  +  x43}8/7+  C\frac{{21}}{{32}}\,\,{\left\{ {1\,\, + \,\,\sqrt[3]{{{x^4}}}} \right\}^{8/7}} + \,\,C
Option A is correct because the integral is solved using substitution, resulting in the given expression.
33 Medium
33 of 411
Application +4 | -1
The integral of f(x)=3xcos⁡xf(x)=3^x \cos x is given by
View answer and explanation
Correct option: C
3x(sin⁡x+cos⁡log⁡3)(1+(log⁡3)2)\frac{3^x(\sin x+\cos \log 3)}{\left(1+(\log 3)^2\right)}
Option D is correct because the integral of 3xcos⁡x3^x \cos x is 3x(log⁡3sin⁡x+cos⁡x)(1+(log⁡3)2)+C\frac{3^x(\log 3 \sin x+\cos x)}{\left(1+(\log 3)^2\right)}+C after integrating by parts twice.
34 Medium
34 of 411
Application +4 | -1
∫(log⁡x)2dx\int(\log x)^2 d x equals:
View answer and explanation
Correct option: B
x(log⁡x)2−2xlog⁡x+2x+cx(\log x)^2-2 x \log x+2 x+c
Option B (x(log⁡x)2−2xlog⁡x+2x+cx(\log x)^2-2 x \log x+2 x+c) is correct because it is the result of applying integration by parts twice to the given integral.
35 Medium
35 of 411
Application +4 | -1
The value of ∫dx3−2x−x2\int \frac{d x}{3-2 x-x^2} will be
View answer and explanation
Correct option: A
14log⁡(3+x1−x)+c\frac{1}{4} \log \left(\frac{3+x}{1-x}\right)+c
Option A (14log⁡(3+x1−x)+c\frac{1}{4} \log \left(\frac{3+x}{1-x}\right)+c) is correct because it is the result of integrating the given function after completing the square and using partial fractions.
36 Medium
36 of 411
Application +4 | -1
If f(0)=f′(0)=0f(0)=f^{\prime}(0)=0 and f′′(x)=tan⁡2xf^{\prime \prime}(x)=\tan ^2 x then f(x)f(x) is
View answer and explanation
Correct option: A
log⁡sec⁡x−12x2\log \sec x-\frac{1}{2} x^2
Option A (log⁡sec⁡x−12x2\log \sec x-\frac{1}{2} x^2) is correct because integrating f′′(x)=tan⁡2xf''(x) = \tan^2 x twice and applying the initial conditions f(0)=f′(0)=0f(0) = f'(0) = 0 yields this result.
37 Medium
37 of 411
Application +4 | -1
∫x2(a+bx)2dx\int \frac{x^2}{(a+b x)^2} d x equals to:
View answer and explanation
Correct option: B
1b3(a+bx−2aln⁡∣a+bx∣−a2a+bx)+C\frac{1}{b^3}\left(a+b x-2 a \ln |a+b x|-\frac{a^2}{a+b x}\right)+C
Option B (1b3(a+bx−2aln⁡∣a+bx∣−a2a+bx)+C\frac{1}{b^3}\left(a+b x-2 a \ln |a+b x|-\frac{a^2}{a+b x}\right)+C) is correct because it is the result of correctly integrating the given expression using substitution and partial fractions.
38 Medium
38 of 411
Analyzing +4 | -1
Let ∫  x21−x  dx=p  (1−x)  (3x2+4x+8)\int \,\,\frac{{{x^2}}}{{\sqrt {1 - x} }}\,\,dx = p\,\,\sqrt {\left( {1 - x} \right)} \,\,\left( {3{x^2} + 4x + 8} \right), then value of pp is
View answer and explanation
Correct option: A
− 215\frac{{ - \,2}}{{15}}
Option A (− 215\frac{{ - \,2}}{{15}}) is correct because substituting 1−x=t21-x = t^2 and integrating yields the correct value for p.
39 Medium
39 of 411
Application +4 | -1
∫dx(x+1)2/3(x−1)4/3\int \frac{d x}{(x+1)^{2 / 3}(x-1)^{4 / 3}} equals
View answer and explanation
Correct option: A
32(1+x1−x)1/3+c\frac{3}{2}\left(\frac{1+x}{1-x}\right)^{1 / 3}+c
Option A is correct because the integral is solved by substitution, where (x+1)/(x-1) = t3, simplifying the integral to a power rule.
40 Medium
40 of 411
Analyzing +4 | -1
If ∫  sin⁡8  x  −  cos⁡8  x1−2  sin⁡2  x    cos⁡2  x    dx=A  sin⁡   2x+B\int \,\,\frac{{{{\sin }^8}\,\,x\,\, - \,\,{{\cos }^8}\,\,x}}{{1 - 2\,\,{{\sin }^2}\,\,x\;\,\,{{\cos }^2}\,\,x}}\,\,\;dx = A\,\,\sin \,\,\,2x + B, then
View answer and explanation
Correct option: A
A=−  12A = - \,\,\frac{1}{2}
Option A is correct because after simplifying the integral, the coefficient of sin⁡ 2x\sin \,2x is − 12- \,\frac{1}{2}.
41 Medium
41 of 411
Application +4 | -1
∫02(x−log⁡2  a)dx=2  log⁡2  (2a)\int \limits_0^2 \left( {x - {{\log }_2}\,\,a} \right)dx = 2\,\,{\log _2}\,\,\left( {\frac{2}{a}} \right), if
View answer and explanation
Correct option: A
a=2a = 2
Option A (a=2a = 2) is correct because substituting a=2a=2 into the integral equation satisfies the given condition.
42 Medium
42 of 411
Application +4 | -1
If ∫sin⁡−1  (2x1+x2)  dx=f(x)  −  log⁡  (1+x2)  +  c\int {\sin ^{ - 1}}\,\,\left( {\frac{{2x}}{{1 + {x^2}}}} \right)\,\,dx = f\left( x \right)\,\, - \,\,\log \,\,\left( {1 + {x^2}} \right)\,\, + \,\,c, then f(x)f\left( x \right) is equal to
View answer and explanation
Correct option: A
2x  tan⁡−1  x2x\,\,{\tan ^{ - 1}}\,\,x
Option A (2x tan⁡−1 x2x\,{\tan ^{ - 1}}\,x) is correct because it is derived by using trigonometric substitution and integration by parts.
43 Medium
43 of 411
Application +4 | -1
If I=∫dxex+4e−x=f(x)+cI=\int \frac{d x}{e^x+4 e^{-x}}=f(x)+c, then f(x)f(x) is equal to
View answer and explanation
Correct option: B
12tan⁡−1(ex2)\frac{1}{2} \tan ^{-1}\left(\frac{e^x}{2}\right)
Option B (12tan⁡−1(ex2)\frac{1}{2} \tan ^{-1}\left(\frac{e^x}{2}\right)) is correct because the integral can be solved using a simple substitution, resulting in the given expression.
44 Medium
44 of 411
Application +4 | -1
∫x+sin⁡x1+cos⁡xdx\int \frac{x+\sin x}{1+\cos x} d x is equal to
View answer and explanation
Correct option: B
xtan⁡x2+cx \tan \frac{x}{2}+c
Option B (xtan⁡x2+cx \tan \frac{x}{2}+c) is correct because it is the result of integrating the given expression using trigonometric identities and integration by parts.
45 Medium
45 of 411
Application +4 | -1
The value of ∫dx(x−a)(b−x)\int \frac{d x}{\sqrt{(x-a)(b-x)}}, is
View answer and explanation
Correct option: A
2sin⁡−1x−ab−a+C2 \sin ^{-1} \sqrt{\frac{x-a}{b-a}}+C
Option A (2sin⁡−1x−ab−a+C2 \sin ^{-1} \sqrt{\frac{{x-a}}{{b-a}}}+C) is correct because the integral evaluates to this expression after a suitable trigonometric substitution.

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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 Medium
46 of 411
Analyzing +4 | -1
If ∫dx(x2−4)x+1=1klog⁡e∣x+1−3x+1+3∣\int \frac{d x}{\left(x^2-4\right) \sqrt{x+1}}=\frac{1}{k} \log _e\left|\frac{\sqrt{x+1}-\sqrt{3}}{\sqrt{x+1}+\sqrt{3}}\right| −12Tan⁡−1x+1+c-\frac{1}{2} \operatorname{Tan}^{-1} \sqrt{x+1}+c then kk equals
View answer and explanation
Correct option: B
434 \sqrt{3}
Option B (434 \sqrt{3}) is correct because after applying a trigonometric substitution and simplifying, the constant kk is found to be 434 \sqrt{3}.
47 Medium
47 of 411
Application +4 | -1
∫1−x1+xdx=\int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} d x=
View answer and explanation
Correct option: A
−21−x+Cos⁡−1x+x−x2+c-2 \sqrt{1-x}+\operatorname{Cos}^{-1} \sqrt{x}+\sqrt{x-x^2}+c
Option A (−21−x+Cos⁡−1x+x−x2+c-2 \sqrt{1-x}+\operatorname{Cos}^{-1} \sqrt{x}+\sqrt{x-x^2}+c) is correct because it is the result of trigonometric substitution and simplification of the integral.
48 Medium
48 of 411
Application +4 | -1
∫(x4−x)1/4x5dx\int \frac{\left(x^4-x\right)^{1 / 4}}{x^5} d x is equal to
View answer and explanation
Correct option: A
415(1−1x3)5/4+c\frac{4}{15}\left(1-\frac{1}{x^3}\right)^{5 / 4}+c
Option A (415(1−1x3)5/4+c\frac{{4}}{{15}}\left(1-\frac{{1}}{{x^3}}\right)^{{5 / 4}}+c) is correct because it is the result of the integration after making the appropriate substitution.
49 Medium
49 of 411
Application +4 | -1
∫log⁡xx2dx\int \frac{\log x}{x^2} d x is equal to
View answer and explanation
Correct option: D
−(1+log⁡x)x+c-\frac{(1+\log x)}{x}+c
Option D (-(1+log⁡x)x+c\frac{(1+\log x)}{x}+c) is correct because by using integration by parts, we can evaluate the integral.
50 Medium
50 of 411
Analyzing +4 | -1
The anti-derivative of f  (x)=log⁡  (log⁡  x)  +  (log⁡  x)−2f\,\,\left( x \right) = \log \,\,(\log \,\,x)\,\, + \,\,{(\log \,\,x)^{ - 2}} whose graph passes through (e,  e)\left( {e,\,\,e} \right), is
View answer and explanation
Correct option: C
x [log⁡  (log⁡  x)  −  (log⁡  x)−1]  +  2ex\,[\log \,\,(\log \,\,x)\,\, - \,\,{(\log \,\,x)^{ - 1}}]\,\, + \,\,2e
Option C (x [log⁡ (log⁡ x) −  (log⁡ x)−1] + 2ex\,[\log \,(\log \,x)\, - \,\,{(\log \,x)^{ - 1}}]\, + \,2e) is correct because it is the antiderivative of the given function that passes through the point (e, e).
51 Medium
51 of 411
Application +4 | -1
The value of the integral ∫0πxsin⁡x1+cos⁡2xdx\int\limits_{0}^\pi \frac{x \sin x}{1+\cos ^2 x} d x is
View answer and explanation
Correct option: B
π24\frac{\pi^2}{4}
Option B (π24\frac{\pi^2}{4}) is correct because using the property ∫0af(x)dx=∫0af(a−x)dx\int_0^a f(x) dx = \int_0^a f(a-x) dx and then substituting cos⁡x=t\cos x = t, we can evaluate the definite integral to get π24\frac{\pi^2}{4}.
52 Medium
52 of 411
Application +4 | -1
∫ex(1−sin⁡x1−cos⁡x)dx\int e^x\left(\frac{1-\sin x}{1-\cos x}\right) d x is equal to
View answer and explanation
Correct option: B
−excot⁡(x/2)-e^x \cot (x / 2)
Option B is correct because the integral simplifies to −excot⁡(x/2)+c-e^x \cot(x/2) + c by rewriting the integrand and recognizing the form ∫ex[f(x)+f′(x)]dx=exf(x)+c\int e^x [f(x) + f'(x)] dx = e^x f(x) + c.
53 Medium
53 of 411
Application +4 | -1
∫(x+1x)n+5(x2−1x2)dx\int\left(x+\frac{1}{x}\right)^{n+5}\left(\frac{x^2-1}{x^2}\right) d x is equal to :
View answer and explanation
Correct option: A
(x+1x)n+6n+6+c\frac{\left({x}+\frac{1}{{x}}\right)^{{n}+6}}{{n}+6}+{c}
Option A ((x+1x)n+6n+6+c\frac{\left(x+\frac{1}{x}\right)^{n+6}}{n+6}+c) is correct because substituting t=x+1xt = x + \frac{1}{x} simplifies the integral, leading to the given result.
54 Medium
54 of 411
Application +4 | -1
∫sin⁡(log⁡x)dx=\int \sin (\log x) d x=
View answer and explanation
Correct option: C
12x[sin⁡(log⁡x)−cos⁡(log⁡x)]\frac{1}{2} x[\sin (\log x)-\cos (\log x)]
Option C (12x[sin⁡(log⁡x)−cos⁡(log⁡x)]\frac{1}{2} x[\sin (\log x)-\cos (\log x)]) is correct because it is the result of integrating sin⁡(log⁡x)\sin (\log x) using integration by parts twice.
55 Medium
55 of 411
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 ( −  12 x2 + c\ - \,\,\frac{1}{2}\,{x^2}\, + \,c) is correct. Simplifying the expression inside the integral using trigonometric identities leads to the result.
56 Medium
56 of 411
Application +4 | -1
∫sin⁡−1x(1−x2)3/2dx=\int \frac{\sin ^{-1} x}{\left(1-x^2\right)^{3 / 2}} d x=
View answer and explanation
Correct option: A
x1−x2sin⁡−1x+12log⁡(1−x2)+c\frac{x}{\sqrt{1-x^2}} \sin ^{-1} x+\frac{1}{2} \log \left(1-x^2\right)+c
Option A (x1−x2sin⁡−1x+12log⁡(1−x2)+c\frac{x}{\sqrt{1-x^2}} \sin ^{-1} x+\frac{1}{2} \log \left(1-x^2\right)+c) is correct because substituting sin⁡t=x\sin t = x simplifies the integral, and integration by parts yields the correct result.
57 Medium
57 of 411
Understanding +4 | -1
If ∫2x+3x2−5x+6dx=9ln⁡(x−3)−7ln⁡(x−2)+A\int \frac{2 x+3}{x^2-5 x+6} d x=9 \ln (x-3)-7 \ln (x-2)+A then A=A=
View answer and explanation
Correct option: C
Constant
Option C (Constant) is correct because after integrating the given function and comparing it with the provided form, the remaining term is a constant.
58 Medium
58 of 411
Application +4 | -1
∫dx3x3(xn+1)\int \frac{d x^3}{x^3\left(x^n+1\right)} equals:
View answer and explanation
Correct option: A
3nln⁡(xnxn+1)\frac{3}{n} \ln \left(\frac{x^n}{x^n+1}\right)
Option A (3nln⁡(xnxn+1)\frac{3}{n} \ln \left(\frac{x^n}{x^n+1}\right)) is correct because it is the result of the integration after substitution and simplification.
59 Medium
59 of 411
Analyzing +4 | -1
∫x4(1−x)41+x2dx=f(x)−4Tan⁡−1x+c\int \frac{x^4(1-x)^4}{1+x^2} d x=f(x)-4 \operatorname{Tan}^{-1} x+c where f(x)f(x) is
View answer and explanation
Correct option: A
x77−23x6+x5−4x33+4x\frac{x^7}{7}-\frac{2}{3} x^6+x^5-\frac{4 x^3}{3}+4 x
The correct answer is A (x77−23x6+x5−4x33+4x\frac{x^7}{7}-\frac{2}{3} x^6+x^5-\frac{4 x^3}{3}+4 x) because after polynomial long division, the integral is straightforward and the result matches option A.
60 Medium
60 of 411
Application +4 | -1
∫(x4−x)1/4x5dx\int \frac{\left(x^{4}-x\right)^{1 / 4}}{x^{5}} d x is equal to
View answer and explanation
Correct option: A
415(1−1x3)5/4+c\frac{4}{15}\left(1-\frac{1}{x^{3}}\right)^{5 / 4}+c
Option A is correct because the integral is solved by substitution, resulting in the given expression.
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