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JEE • MATHEMATICS • INDEFINITE INTEGRATION

Indefinite integration practice · Medium

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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 391-411 of 411 questions
391 Medium
391 of 411
Application +4 | -1
The value of ∫x7(1−x2)5dx\int \frac{x^7}{\left(1-x^2\right)^5} d x, is
View answer and explanation
Correct option: B
18x8(1−x2)4+C\frac{1}{8} \frac{x^8}{\left(1-x^2\right)^4}+C
Option D (None of these) is correct because the integral evaluates to 18x8(1−x2)4+C\frac{1}{8} \frac{x^8}{(1-x^2)^4}+C, but the explanation provided uses a substitution that changes the variable to theta and then back to x, which is not necessary.
392 Medium
392 of 411
Understanding +4 | -1
If fn(x)=log⁡log⁡…log⁡xf_n(x)=\log \log \ldots \log x (log is repeated nn times) then ∫[xf1(x)f2(x)…fn(x)]−1 dx\int\left[x f_1(x) f_2(x) \ldots f_n(x)\right]^{-1} \mathrm{~d} x is equal to
View answer and explanation
Correct option: A
fn+1(x)+Cf_{n+1}(x)+C
Option A (fn+1(x)+Cf_{n+1}(x)+C) is correct because it is the result of applying the substitution method and integrating.
393 Medium
393 of 411
Application +4 | -1
∫x2+sin⁡2x1+x2sec⁡2xdx\int {\frac{{{x^2} + {{\sin }^2}x}}{{1 + {x^2}}}{{\sec }^2}xdx} is equal to
View answer and explanation
Correct option: B
tan⁡x−tan⁡−1x+c\tan x - {\tan ^{ - 1}}x + c
Option B ((x−1x)K+8k+8+c\frac{\left(x-\frac{1}{x}\right)^{K+8}}{k+8}+c) is correct because after simplifying and substituting t=x−1xt = x - \frac{1}{x}, the integral becomes a simple power rule integral.
394 Medium
394 of 411
Analyzing +4 | -1
∫esec⁡x⋅sec⁡3x(sin⁡2x+cos⁡x+sin⁡x+sin⁡xcos⁡x)dx=\int e^{\sec x} \cdot \sec ^3 x \left(\sin ^2 x+\cos x+\sin x+\sin x \cos x\right) d x=
View answer and explanation
Correct option: A
esec⁡x(sec⁡x+tan⁡x)+e^{\sec x}(\sec x+\tan x)+
The correct answer is A (esec⁡x(sec⁡x+tan⁡x)+ce^{\sec x}(\sec x+\tan x)+c) because the integral simplifies to this expression after applying integration by parts and simplification.
395 Medium
395 of 411
Application +4 | -1
∫xtan⁡−1xdx=\int x \tan ^{-1} x d x=
View answer and explanation
Correct option: A
12(x2+1)tan⁡−1x−12x+c\frac{1}{2}\left(x^2+1\right) \tan ^{-1} x-\frac{1}{2} x+c
Option A is correct because the integral is solved using integration by parts, simplifying the resulting expression to match the given form.
396 Medium
396 of 411
Analyzing +4 | -1
∫  cos⁡  x  −  cos⁡3 x1  −  cos⁡3  x  dx\int \,\,\sqrt {\frac{{\cos \,\,x\,\, - \,\,{{\cos }^3}\,x}}{{1\,\, - \,\,{{\cos }^3}\,\,x}}} \,\,dx is equal to
View answer and explanation
Correct option: C
23  cos⁡−1  (cos⁡3/2  x)+C\frac{2}{3}\,\,{\cos ^{ - 1}}\,\,\left( {{{\cos }^{3/2}}\,\,x} \right) + C
Option C (23,cos⁡−1,(\frac{2}{3}\\,{\cos ^{ - 1}}\\,\left( {{{{}}}} \right.\cos }}^{3/2}}\\,x} \right) + C}}^{3/2}}\\,x} \right) + C) is correct because it represents the result of the integration after simplification and substitution.
397 Medium
397 of 411
Application +4 | -1
If ∫tan⁡7xdx=f(x)+ln⁡∣cos⁡x∣+c\int \tan ^7 x d x=f(x)+\ln |\cos x|+c, then
View answer and explanation
Correct option: D
f(x)f(x) is a polynomial of degree 6 in tan⁡x\tan x
Option D (f(x)f(x) is a polynomial of degree 6 in tan⁡x\tan x) is correct because the integral of tan⁡7x\tan^7 x can be expressed as a polynomial of degree 6 in tan⁡x\tan x plus ln⁡∣cos⁡x∣+c\ln |\cos x| + c.
398 Medium
398 of 411
Application +4 | -1
∫ex(1−x)2(1+x2)2dx=\int e^x \frac{(1-x)^2}{\left(1+x^2\right)^2} d x=
View answer and explanation
Correct option: C
ex(11+x2)+ce^x\left(\frac{1}{1+x^2}\right)+c
Option C (ex(11+x2)+ce^x\left(\frac{1}{1+x^2}\right)+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.
399 Medium
399 of 411
Application +4 | -1
The value of ∫x(xx)x  (2  log⁡  x  +  1)  dx\int x{\left( {{x^x}} \right)^x}\,\,\left( {2\,\,\log \,\,x\,\, + \,\,1} \right)\,\,dx is
View answer and explanation
Correct option: A
(xx)x  +  c{\left( {{x^x}} \right)^x}\,\, + \,\,c
Option A ((xx)x+c{\left( {{x^x}} \right)^x}\\ + \\ c) is correct because the integral is solved by recognizing the derivative of (x^x)^x and using substitution.
400 Medium
400 of 411
Analyzing +4 | -1
Integration of 1x2+  9\frac{1}{{\sqrt {{x^2} + \,\,9} }} with respect to (x2+  1)\left( {{x^2} + \,\,1} \right) is equal to
View answer and explanation
Correct option: C
2 x2+  9  +  C2\,\sqrt {{x^2} + \,\,9} \,\, + \,\,C
The correct answer is C (2,x2+  9  +  C2\\,\sqrt {{{{x^2} + \,\,9}}} \,\, + \,\,C) because the integral is solved by recognizing the differential and integrating.
401 Medium
401 of 411
Analyzing +4 | -1
∫ex2(1−x2)(1−x2)1−2x2dx\int \frac{e^{x \sqrt{2}}\left(1-x^2\right)}{(1-x \sqrt{2}) \sqrt{1-2 x^2}} d x is equal to
View answer and explanation
Correct option: C
122ex21+x21−x2+c\frac{1}{2 \sqrt{2}} e^{x \sqrt{2}} \sqrt{\frac{1+x \sqrt{2}}{1-x \sqrt{2}}}+c
The correct answer is C (122ex21+x21−x2+c\frac{1}{2 \sqrt{2}} e^{x \sqrt{2}} \sqrt{\frac{1+x \sqrt{2}}{1-x \sqrt{2}}}+c) because after substitution and simplification, the integral can be solved, and the result matches option C.
402 Medium
402 of 411
Application +4 | -1
Let ∫ex(f(x)−f′(x))dx=ϕ(x)\int e^x\left(f(x)-f^{\prime}(x)\right) d x=\phi(x), then ∫exf(x)dx\int e^x f(x) d x is equal to
View answer and explanation
Correct option: C
12{ϕ(x)+ext(x)}+c\frac{1}{2}\left\{\phi(x)+e^x t(x)\right\}+c
Option C (12{ϕ(x)+ext(x)}\frac{1}{2}\{\phi(x)+e^x t(x)\}+c) is correct because by integrating ex(f(x)+f′(x))e^x(f(x)+f'(x)) and ex(f(x)−f′(x))e^x(f(x)-f'(x)) and adding them, we can isolate the integral of exf(x)e^x f(x).
403 Medium
403 of 411
Analyzing +4 | -1
Given, f(x)=∣0x2−sin⁡xcos⁡x−2sin⁡x−x201−2x2−cos⁡x2x−10∣f(x)=\left|\begin{array}{ccc}0 & x^2-\sin x & \cos x-2 \\ \sin x-x^2 & 0 & 1-2 x \\ 2-\cos x & 2 x-1 & 0\end{array}\right|, then ∫f(x)dx\int f(x) d x is equal to
View answer and explanation
Correct option: D
None of the above
The correct answer is D (None of the above) because the determinant is zero, making the integral zero as well.
404 Medium
404 of 411
Application +4 | -1
Evaluate: ∫sin⁡(x−a)sin⁡(x+a)dx\int \frac{\sin (x-a)}{\sin (x+a)} d x
View answer and explanation
Correct option: C
(x+a)cos⁡2a−sin⁡2alog⁡∣sin⁡(x+a)∣+c(x+a) \cos 2 a-\sin 2 a \log |\sin (x+a)|+c
Option C ((x+a)cos⁡2a−sin⁡2alog⁡∣sin⁡(x+a)∣+c(x+a) \cos 2 a-\sin 2 a \log |\sin (x+a)|+c) is correct because the integral is solved by substituting x+a=tx+a=t and using trigonometric identities.
405 Medium
405 of 411
Application +4 | -1
∫  dx(a2+x2)3/2\int \,\,\frac{{dx}}{{{{\left( {{a^2} + {x^2}} \right)}^{3/2}}}} is equal to
View answer and explanation
Correct option: B
xa2(a2+x2)1/2+ c\frac{x}{{{a^2}{{\left( {{a^2} + {x^2}} \right)}^{1/2}}}} + \,c
Option B is correct because after trigonometric substitution and simplification, the integral evaluates to xa2(a2+x2)1/2+ c\frac{x}{{{{a^2}{{\left( {{a^2} + {x^2}} \right)}^{1/2}}}}} + \,c.
406 Medium
406 of 411
Application +4 | -1
∫  x1  +  x34  dx\int \,\,\frac{{\sqrt x }}{{1\,\, + \,\,\sqrt[4]{{{x^3}}}}}\,\,dx is equal to
View answer and explanation
Correct option: B
43  [1+x3/4  −  log⁡e  (1  +  x3/4)]+C\frac{4}{3}\,\,\left[ {1 + {x^{3/4}}\,\, - \,\,{{\log }_e}\,\,\left( {1\,\, + \,\,{x^{3/4}}} \right)} \right] + C
Option B (43  [1+x3/4  −  log⁡e  (1  +  x3/4)]+C\frac{4}{3}\,\,\left[ {1 + {x^{3/4}}\,\, - \,\,{{\log }_e}\,\,\left( {1\,\, + \,\,{x^{3/4}}} \right)} \right] + C) is correct because after substitution and integration, the result matches this option.
407 Medium
407 of 411
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
∣x∣nxn+1+C\frac{|x|^n x}{n+1}+C
Option C (∣x∣nxn+1+C\frac{|x|^n x}{n+1}+C) is correct because it accounts for both positive and negative values of xx when nn is an odd positive integer.
408 Medium
408 of 411
Application +4 | -1
If ∫dxx−x3=Aln⁡∣x21−x2∣+c\int \frac{d x}{x-x^3}=A \ln \left|\frac{x^2}{1-x^2}\right|+c, then AA is equal to
View answer and explanation
Correct option: B
1/21/2
Option B (1/2) is correct because the integral simplifies to the given form with A = 1/2 after partial fraction decomposition.
409 Medium
409 of 411
Application +4 | -1
∫dxxlog⁡xlog⁡(log⁡x)=\int \frac{d x}{x \log x \log (\log x)}=
View answer and explanation
Correct option: B
log⁡[log⁡(log⁡x)]+c\log [\log (\log x)]+c
Option B (log⁡[log⁡(log⁡x)]+c\log [\log (\log x)]+c) is correct because it is the result of two u-substitutions.
410 Medium
410 of 411
Application +4 | -1
∫dxcos⁡  x  −  sin⁡  x\int \frac{{dx}}{{\cos \,\,x\,\, - \,\,\sin \,\,x}} is equal to
View answer and explanation
Correct option: D
12  log⁡  ∣ tan⁡  (x2  +  3π8)+c ∣\frac{1}{{\sqrt 2 }}\,\,\log \,\,\left| {\,\tan \,\,\left( {\frac{x}{2}\,\, + \,\,\frac{{3\pi }}{8}} \right) + c\,} \right|
Option D is correct. The integral ∫dxcos⁡  x  −  sin⁡  x\int \frac{{dx}}{{\cos \,\,x\,\, - \,\,\sin \,\,x}} can be solved by rewriting the denominator and using trigonometric identities.
411 Medium
411 of 411
Analyzing +4 | -1
If
∫tan⁡x1+tan⁡x+tan⁡2xdx=x−KAtan⁡−1(Ktan⁡x+1A)+C,\int \frac{\tan x}{1+\tan x+\tan ^2 x} d x=x-\frac{K}{\sqrt{A}} \tan ^{-1}\left(\frac{K \tan x+1}{\sqrt{A}}\right)+C,
( CC is a constant of integration), then the ordered pair (K,A)(K, A) is equal to
View answer and explanation
Correct option: A
(2,3)(2,3)
Option A ((2,3)) is correct because after solving the integral and comparing with the given form, we find K=2 and A=3.
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