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JEE • MATHEMATICS • INDEFINITE INTEGRATION

Indefinite integration practice · Mixed difficulty

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361 Medium
361 of 995
Analyzing +4 | -1
If ∫esec⁡x(sec⁡xtan⁡xf(x)+(secxtan⁡x+sec⁡2x))dx=esec⁡xf(x)+C\int {{e^{\sec x}}} (\sec x\tan x\mathrm{ }f(x) + (sec\mathrm{ }x\tan x + {\sec ^2}x))\mathrm{ }dx = {e^{\sec x}}f(x) + C, then a possible choice of f(x)f(x) is
View answer and explanation
Correct option: B
sec⁡x+tan⁡x+12\sec x + \tan x + \frac{1}{2}
Option B (sec⁡x+tan⁡x+12\sec x + \tan x + \frac{1}{2}) is correct because differentiating esec⁡xf(x)e^{\sec x}f(x) and comparing with the integrand allows us to identify f(x)f(x).
362 Medium
362 of 995
Application +4 | -1
If ∫ex(1+sin⁡x)dx1+cos⁡x=exf(x)+c\int \frac{e^x(1+\sin x) d x}{1+\cos x}=e^x f(x)+c, then f(x)=f(x)=
View answer and explanation
Correct option: C
tan⁡x2\tan \frac{x}{2}
Option C (tan⁡x2\tan \frac{x}{2}) is correct because when the integral is simplified, the function f(x) is identified as tan⁡x2\tan \frac{x}{2} based on the form ∫ex[f(x)+f′(x)]dx=exf(x)+c\int e^x[f(x) + f'(x)] dx = e^x f(x) + c.
363 Medium
363 of 995
Analyzing +4 | -1
If ∫3tan⁡(x−π4)cos⁡2xtan⁡3x+tan⁡2x+tan⁡xdx=k  tan⁡−1(tan⁡x+1+cot⁡x)+C,\int {\frac{{3\tan \left( {x - \frac{\pi }{4}} \right)}}{{{{\cos }^2}x\sqrt {{{\tan }^3}x + {{\tan }^2}x + \tan x} }}dx = k\,\,{{\tan }^{ - 1}}\left( {\sqrt {\tan x + 1 + \cot x} } \right) + C,} then the value of kk is: [where CC is constant of integration].
View answer and explanation
Correct option: C
6
Option C, 6, is correct because after substitution and integration, the value of k is found to be 6.
364 Medium
364 of 995
Application +4 | -1
∫dx1−tan⁡2x=1λsin⁡−1(λsin⁡x)+C\int \frac{d x}{\sqrt{1-\tan ^2 x}}=\frac{1}{\lambda} \sin ^{-1}(\lambda \sin x)+C, then λ=\lambda=
View answer and explanation
Correct option: A
2\sqrt{2}
Option A (2\sqrt{2}) is correct because by manipulating the integrand and using a trigonometric substitution, the integral can be solved, revealing that λ=2\lambda = \sqrt{2}.
365 Hard
365 of 995
Analyzing +4 | -1
∫−11  x3  +  ∣ x ∣  +  1x2  +  2∣ x ∣  + 1  dx\int \limits_{ - 1}^1 \,\,\frac{{{x^3}\,\, + \,\,\left| {\,x\,} \right|\,\, + \,\,1}}{{{x^2}\,\, + \,\,2\left| {\,x\,} \right|\,\, + \,1}}\,\,dx is equal to
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the integral evaluates to 2ln2, which is not among the given options.
366 Medium
366 of 995
Application +4 | -1
∫ln⁡(x+x2+1)x2+1dx\int \frac{\ln \left(x+\sqrt{x^2+1}\right)}{\sqrt{x^2+1}} d x is equal to
View answer and explanation
Correct option: C
12ln⁡2(x+x2+1)+c\frac{1}{2} \ln ^2\left(x+\sqrt{x^2+1}\right)+c
Option C (12ln⁡2(x+x2+1)+c\frac{1}{2} \ln ^2(x+\sqrt{x^2+1})+c) is correct because it is the result of integrating the given expression using substitution.
367 Easy
367 of 995
Understanding +4 | -1
If ∫dxsin⁡xcos⁡x=log⁡∣f(x)∣+C\int \frac{\mathrm{d} x}{\sin x \cos x}=\log |f(x)|+C then
View answer and explanation
Correct option: B
f(x)=tan⁡xf(x)=\tan x
Option B (f(x)=tan⁡xf(x)=\tan x) is correct because the integral evaluates to log⁡∣tan⁡x∣+C\log |\tan x| + C.
368 Easy
368 of 995
Application +4 | -1
∫sin⁡−1x(1−x2)1/2dx=\int \frac{\sin ^{-1} x}{\left(1-x^2\right)^{1 / 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 (1-x^2) + c) is correct because the integral can be solved by substitution and then integration by parts.
369 Medium
369 of 995
Analyzing +4 | -1
The integral ∫(1−13)(cos⁡x−sin⁡x)(1+23sin⁡2x)dx\int \frac{\left(1-\frac{1}{\sqrt{3}}\right)(\cos x-\sin x)}{\left(1+\frac{2}{\sqrt{3}} \sin 2 x\right)} d x is equal to
View answer and explanation
Correct option: A
12log⁡e∣tan⁡(x2+π12)tan⁡(x2+π6)∣+C\frac{1}{2} \log _e\left|\frac{\tan \left(\frac{x}{2}+\frac{\pi}{12}\right)}{\tan \left(\frac{x}{2}+\frac{\pi}{6}\right)}\right|+C
Option A (12log⁡e∣tan⁡(x2+π12)tan⁡(x2+π6)∣+C\frac{{1}}{{2}} \log _e\left| {\frac{{\tan \left( {\frac{{x}}{{2}} + \frac{{\pi }}{{12}}} \right)}}{{\tan \left( {\frac{{x}}{{2}} + \frac{{\pi }}{{6}}} \right)}}} \right| + C) is correct because after simplification and integration, the given integral evaluates to this expression.
370 Medium
370 of 995
Application +4 | -1
∫xtan⁡xsec⁡2xdx=\int x \tan x \sec ^2 x d x=
View answer and explanation
Correct option: B
xsec⁡2x2−12tan⁡x+c\frac{x \sec ^2 x}{2}-\frac{1}{2} \tan x+c
Option B (xsec⁡2x2−12tan⁡x+c\frac{x \sec ^2 x}{2}-\frac{1}{2} \tan x+c) is correct because it is the result of applying integration by parts to the given integral.
371 Medium
371 of 995
Application +4 | -1
∫f(x)g′(x)+f′(x)g(x)f(x) g(x){log⁡f(x)+log⁡g(x)}dx\int \frac{f(x) g^{\prime}(x)+f^{\prime}(x) g(x)}{f(x) \,g(x)}\{\log f(x)+\log g(x)\} d x is equal to
View answer and explanation
Correct option: B
12[log⁡{f(x) g(x)}]2+C\frac{1}{2}[\log \{f(x)\, g(x)\}]^2+C
Option B is correct because the integral simplifies to a form that can be solved by recognizing the derivative of log(f(x)g(x)).
372 Medium
372 of 995
Analyzing +4 | -1
If ∫  uv′′ dx=uv′−  vu′+a\int \,\,uv''\,dx = uv' - \,\,vu' + a, then aa is equal to
View answer and explanation
Correct option: A
∫u′′v    dx\int u''v\;\,\,dx
Option A is correct because after applying integration by parts twice, aa is equal to ∫u′′v    dx\int u''v\;\,\,dx.
373 Easy
373 of 995
Application +4 | -1
∫exsin⁡x(sin⁡x+2cos⁡x)dx=\int e^x \sin x(\sin x+2 \cos x) d x=
View answer and explanation
Correct option: A
exsin⁡2x+ce^x \sin ^2 x+c
Option A is correct because the integral simplifies to ∫ex(sin⁡2x+sin⁡2x)dx\int e^x(\sin^2 x + \sin 2x) dx, which directly integrates to exsin⁡2x+ce^x \sin^2 x + c.
374 Easy
374 of 995
Understanding +4 | -1
The primitive of the function f(x)=(1−1x2)ax+1x,x>0,f(x)=\left(1-\frac{1}{x^2}\right) a^{x+\frac{1}{x}}, x > 0, is
View answer and explanation
Correct option: A
ax+1xlog⁡ea\frac{a^{x+\frac{1}{x}}}{\log _e a}
Option A is correct because the primitive (indefinite integral) of the given function is found by recognizing the derivative pattern and applying the appropriate integration formula for exponential functions.
375 Medium
375 of 995
Application +4 | -1
∫ex(x2+1)(x+1)2dx=\int e^x \frac{\left(x^2+1\right)}{(x+1)^2} d x=
View answer and explanation
Correct option: D
ex(x−1x+1)+ce^x\left(\frac{x-1}{x+1}\right)+c
Option D (ex(x−1x+1)+ce^x\left(\frac{x-1}{x+1}\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.

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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
376 Easy
376 of 995
Application +4 | -1
∫  x(x2−a2)  (x2−b2)  dx\int \,\,\frac{x}{{\left( {{x^2} - {a^2}} \right)\,\,\left( {{x^2} - {b^2}} \right)}}\,\,dx is equal to
View answer and explanation
Correct option: C
12 (a2−  b2)  log⁡  ∣ x2−a2x2−b2 ∣  +  c\frac{1}{{2\,({a^2} - \,\,{b^2})}}\,\,\log \,\,\left| {\,\frac{{{x^2} - {a^2}}}{{{x^2} - {b^2}}}\,} \right|\,\, + \,\,c
Option C (12 (a2− b2)  log⁡  ∣ x2− a2x2− b2 ∣  +  c\frac{1}{{2\,({a^2} - \,{b^2})}}\,\,\log \,\,\left| {\,\frac{{{{x^2} - \,{a^2}}}}{{{{x^2} - \,{b^2}}}}}\,\right|\,\, + \,\,c) is correct because it represents the correct partial fraction decomposition and integration of the given expression.
377 Easy
377 of 995
Application +4 | -1
∫  exlog⁡a  ⋅  ex   dx\int \,\,{e^{x\log a}}\,\, \cdot \,\,{e^x}\;\,dx is equal to
View answer and explanation
Correct option: B
(ae)xlog⁡  (ae)\frac{{{{\left( {ae} \right)}^x}}}{{\log \,\,\left( {ae} \right)}}
Option B ((ae)xlog⁡  (ae)\frac{{{{\left( {ae} \right)}^x}}}{{\log \,\,\left( {ae} \right)}}) is correct because after simplifying the integrand and integrating, the result matches this option.
378 Easy
378 of 995
Application +4 | -1
If f(3x−43x+4)=x+2f\left(\frac{3 x-4}{3 x+4}\right)=x+2, then ∫f(x)dx\int f(x) d x is equal to
View answer and explanation
Correct option: B
−83log⁡e∣1−x∣+23x+C-\frac{8}{3} \log _e|1-x|+\frac{2}{3} x+C
Option B (−83log⁡e∣1−x∣+23x+C\frac{{-8}}{{3}} \log _e|1-x|+\frac{{2}}{{3}} x+C) is correct because it is the result of integrating f(x) after finding the expression for f(x).
379 Medium
379 of 995
Application +4 | -1
∫sin⁡2xsin⁡4x+cos⁡4xdx=\int \frac{\sin 2 x}{\sin ^4 x+\cos ^4 x} d x=
View answer and explanation
Correct option: B
tan⁡−1(tan⁡2x)+c\tan ^{-1}\left(\tan ^2 x\right)+c
Option B is correct because the integral can be solved by rewriting the integrand using trigonometric identities and then applying substitution.
380 Medium
380 of 995
Application +4 | -1
If ∫dx1+tan⁡x=px+qlog⁡e∣cos⁡x+sin⁡x∣+c\int \frac{d x}{1+\tan x}=p x+q \log _e|\cos x+\sin x|+c then p+qp+q equals
View answer and explanation
Correct option: D
1
Option D (1) is correct because it is the sum of p and q, which are calculated using the formula for integrals of the form ∫dx1+tan⁡x\int \frac{{dx}}{{1 + \tan x}}.
381 Easy
381 of 995
Application +4 | -1
If f(x)=1cos⁡2x1+tan⁡xf(x)=\frac{1}{\cos ^2 x \sqrt{1+\tan x}} then its antiderivative F(x)F(x) (given that F(0)=4F(0)=4 ) is
View answer and explanation
Correct option: C
2(1+tan⁡x+1)2(\sqrt{1+\tan x}+1)
Option C (2(1+tan⁡x+1)2(\sqrt{{1+\tan x}}+1)) is correct because it represents the antiderivative of the given function with the correct constant of integration.
382 Medium
382 of 995
Application +4 | -1
∫1{(x−1)3  (x+2)5}1/4dx\int \frac{1}{{{{\left\{ {{{\left( {x - 1} \right)}^3}\,\,{{\left( {x + 2} \right)}^5}} \right\}}^{1/4}}}}dx is equal to
View answer and explanation
Correct option: A
43  (x  −  1x  +  2)1/4+  C\frac{4}{3}\,\,{\left( {\frac{{x\,\, - \,\,1}}{{x\,\, + \,\,2}}} \right)^{1/4}} + \,\,C
Option A is correct because the integral simplifies to a form where the power rule can be applied after a suitable substitution.
383 Medium
383 of 995
Application +4 | -1
∫cot⁡−1(1−x+x2−x3+x4−x5+…)dx(∣x∣<1)\int \cot ^{-1}\left(1-x+x^2-x^3+x^4-x^5+\ldots\right) d x(|x| < 1)
View answer and explanation
Correct option: A
(x+1)tan⁡−1(x+1)−log⁡1+(x+1)2+C(x+1) \tan ^{-1}(x+1)-\log \sqrt{1+(x+1)^2}+C
Option A ((x+1)tan⁡−1(x+1)−log⁡1+(x+1)2+C(x+1) \tan ^{-1}(x+1)-\log \sqrt{1+(x+1)^2}+C) is correct because it is the result of integrating the given expression using integration by parts and simplifying the result.
384 Easy
384 of 995
Application +4 | -1
∫(xx)x(2xlog⁡ex+x)dx\int\left(x^x\right)^x\left(2 x \log _e x+x\right) d x is equal to
View answer and explanation
Correct option: B
(xx)x+C\left(x^x\right)^x+C
Option B is correct because the integral simplifies to ∫d(xx2)\int d(x^{x^2}), which evaluates to (xx)x+C(x^x)^x + C.
385 Easy
385 of 995
Application +4 | -1
∫  ex(2x  −  2x2)dx\int \,\,{e^x}\left( {\frac{2}{x}\,\, - \,\,\frac{2}{{{x^2}}}} \right)dx is equal to
View answer and explanation
Correct option: C
2exx  +  c\frac{{2{e^x}}}{x}\,\, + \,\,c
Option C (2exx+c\frac{{2{e^x}}}{x} + c) is correct because the integral is of the form ∫ex[f(x)+f′(x)]dx=exf(x)+c\int e^x [f(x) + f'(x)] dx = e^x f(x) + c.
386 Easy
386 of 995
Application +4 | -1
∫sin⁡3xcos⁡2xdx=\int \sin ^3 x \cos ^2 x d x=
View answer and explanation
Correct option: A
cos⁡5x5−cos⁡3x3+c\frac{\cos ^5 x}{5}-\frac{\cos ^3 x}{3}+c
Option A is correct because the integral of sin⁡3xcos⁡2x\sin^3 x \cos^2 x can be solved by substituting t=cos⁡xt = \cos x and simplifying the resulting expression.
387 Medium
387 of 995
Application +4 | -1
If u=∫eaxsin⁡bx dxu=\int e^{a x} \sin b x \,d x and v=∫eaxcos⁡bx dx,v=\int e^{a x} \cos b x \,d x, then (u2+v2)(a2+b2)=\left(u^2+v^2\right)\left(a^2+b^2\right)=
View answer and explanation
Correct option: B
e2axe^{2 a x}
Option B, e2axe^{2 a x}, is correct because when uu and vv are squared and added, and then multiplied by (a2+b2)(a^2+b^2), the result simplifies to e2axe^{2ax}.
388 Easy
388 of 995
Application +4 | -1
If ∫4x2−4x+3dx=12log⁡e(2x−1+4x2−4x+3)+f(x)+c\int \sqrt{4 x^2-4 x+3} d x=\frac{1}{2} \log _e\left(2 x-1+\sqrt{4 x^2-4 x+3}\right)+f(x)+c and then f(x)f(x) equals
View answer and explanation
Correct option: A
2x−144x2−4x+3\frac{2 x-1}{4} \sqrt{4 x^2-4 x+3}
Option A is correct because the integral is solved by completing the square and using a standard integral formula, resulting in f(x) = (2x-1)/4 * sqrt(4x2-4x+3).
389 Medium
389 of 995
Application +4 | -1
∫  x+sin⁡  x1+cos⁡  x  dx\int \,\,\frac{{x + \sin \,\,x}}{{1 + \cos \,\,x}}\,\,dx is equal to
View answer and explanation
Correct option: A
x  tan⁡  x2+Cx\,\,\tan \,\,\frac{x}{2} + C
Option A is correct because the integral simplifies to x  tan⁡  x2+Cx\,\,\tan \,\,\frac{x}{2} + C after applying integration by parts.
390 Easy
390 of 995
Application +4 | -1
The value of ∫sin⁡x3dx\int \sin \sqrt[3]{x} d x is
View answer and explanation
Correct option: B
3{(2−x2/3)cos⁡x1/3+2x1/3sin⁡x1/3}+C3\left\{\left(2-x^{2 / 3}\right) \cos x^{1 / 3}+2 x^{1 / 3} \sin x^{1 / 3}\right\}+C
Option D is correct because after substituting x1/3=θx^{1/3} = \theta and integrating by parts twice, we obtain the given expression.
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