Skip to main content
Back to INDEFINITE INTEGRATION

Your Progress

0 / 995 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • MATHEMATICS • INDEFINITE INTEGRATION

Indefinite integration practice · Mixed difficulty

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

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.

Take a chapter quiz →
Showing 391-420 of 995 questions

Preparing practice and saved progress…

391 Easy
391 of 995
Analyzing +4 | -1
∫{(log⁡x−1)1+(log⁡x)2}2dx\int\left\{\frac{(\log x-1)}{1+(\log x)^2}\right\}^2 d x is equal to
View answer and explanation
Correct option: B
x(log⁡x)2+1+c\frac{x}{(\log x)^2+1}+c
Option B (x(log⁡x)2+1+c\frac{x}{(\log x)^2+1}+c) is correct because it is the result of integrating the given expression using substitution and integration by parts.
392 Easy
392 of 995
Analyzing +4 | -1
If ∫cos⁡ x dxsin⁡3x(1+sin⁡6)x)2/3=f(x)(1+sin⁡6x)1/λ+c\int {\frac{{\cos \,x\,dx}}{{{{\sin }^3}x\left( {1 + {{\sin }^6}} \right)x{)^{2/3}}}} = f\left( x \right){{\left( {1 + {{\sin }^6}x} \right)}^{1/\lambda }} + c} where cc is a constant of integration, then is equal to
View answer and explanation
Correct option: C
-2
Option C (-2) is correct because by substituting t=1sin⁡6x+1=t3t = \frac{1}{{{\sin }^6}x} + 1 = {t^3} and solving the integral, we can find the value of λf(π3)\lambda f(\frac{\pi}{3}).
393 Easy
393 of 995
Application +4 | -1
∫1+cot⁡2x1+cot⁡xdx=\int \frac{1+\cot ^2 x}{1+\cot x} d x=
View answer and explanation
Correct option: A
−log⁡(1+cot⁡x)+c-\log (1+\cot x)+c
Option A (−log⁡(1+cot⁡x)+c- \log (1+\cot x)+c) is correct because the integral simplifies to this form after substitution.
394 Medium
394 of 995
Application +4 | -1
∫11+cos⁡2xdx=\int \frac{1}{1+\cos ^2 x} d x=
View answer and explanation
Correct option: C
12tan⁡−1(12tan⁡x)+c\frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{1}{\sqrt{2}} \tan x\right)+c
Option C is correct because the integral is solved by dividing by cos²x and using a substitution, resulting in 12tan⁡−1(12tan⁡x)+c\frac{1}{\sqrt{2}} \tan^{-1}(\frac{1}{\sqrt{2}} \tan x)+c.
395 Easy
395 of 995
Application +4 | -1
∫1+x21−x2dx=\int \frac{1+x^2}{\sqrt{1-x^2}} d x=
View answer and explanation
Correct option: A
32sin⁡−1x−12x1−x2+c\frac{3}{2} \sin ^{-1} x-\frac{1}{2} x \sqrt{1-x^2}+c
Option A is correct because the integral evaluates to 32sin⁡−1x−12x1−x2+c\frac{3}{2} \sin^{-1} x - \frac{1}{2}x \sqrt{1-x^2} + c after trigonometric substitution and simplification.
396 Easy
396 of 995
Understanding +4 | -1
If ∫dxa2−x2=sin⁡−1x3+c\int \frac{d x}{\sqrt{a^2-x^2}}=\sin ^{-1} \frac{x}{3}+c, then a=a=
View answer and explanation
Correct option: A
3
3 is the correct answer because the integral of 1/sqrt(a2 - x2) is arcsin(x/a) + c, and comparing it to the given integral, a must be 3.
397 Medium
397 of 995
Analyzing +4 | -1
If ∫f(x)cos⁡  x dx=12  {f(x)}2+  c,   then    f(x)\int f\left( x \right)\cos \,\,x\,dx = \frac{1}{2}\,\,{\left\{ {f\left( x \right)} \right\}^2} + \,\,c,\;\,\mathrm{then}\,\,\;f\left( x \right) is

View answer and explanation
Correct option: B
sin⁡  x+c\sin \,\,x + c
Option B (sin⁡  x+c\sin \,\,x + c) is correct because the derivative of 12{f(x)}2+c\frac{1}{2}\{f(x)\} ^2 + c leads to f′(x)=cos⁡xf'(x) = \cos x, and integrating gives f(x)=sin⁡x+cf(x) = \sin x + c.
398 Easy
398 of 995
Application +4 | -1
∫  dxx  (log⁡   x)  (log⁡  log⁡  x)  …  (log⁡  log⁡.... x)⏟8  times\int \,\,\frac{{dx}}{{x\,\,(\log \,\,\,x)\,\,(\log \,\,\log \,\,x)\,\, \ldots \,\,\underbrace {(\log \,\,\log ....\,x)}_{8\,\,\mathrm{times}}}} is equal to
View answer and explanation
Correct option: C
(log⁡  log⁡  ....  x)⏟9  times  +  c\underbrace {(\log \,\,\log \,\,....\,\,x)}_{9\,\,\mathrm{times}}\,\, + \,\,c
Option C is correct because after repeated substitution, the integral simplifies to log(log...x) nine times + c.
399 Hard
399 of 995
Analyzing +4 | -1
∫  x2  −  1x (x2  +  α  x  +  1)  (x2  +  β  x  +  1)  dx\int \,\,\frac{{{x^2}\,\, - \,\,1}}{{x\,\sqrt {\left( {{x^2}\,\, + \,\,\alpha \;x\,\, + \,\,1} \right)\,\,\left( {{x^2}\,\, + \,\,\beta \;x\,\, + \,\,1} \right)} }}\,\,dx is equal to
View answer and explanation
Correct option: A
log⁡  { x2+  α  x  +1  +  x2+  β  x  +1x }  +  C\log \,\,\left\{ {\,\frac{{\sqrt {{x^2} + \,\,\alpha \;x\,\, + 1} \,\, + \,\,\sqrt {{x^2} + \,\,\beta \;x\,\, + 1} }}{{\sqrt x }}\,} \right\}\,\, + \,\,C
Option A (log⁡  { x2+  α  x  +1  +  x2+  β  x  +1x}}  +  C\log \,\,\left\{ {\,\frac{{\sqrt {{{{x^2}} + \,\,\alpha \;x\,\, + 1}} \,\, + \,\,\sqrt {{{{x^2}} + \,\,\beta \;x\,\, + 1}} }}{{\sqrt x }}}\} \right\}\,\, + \,\,C) is correct because the integral is solved by dividing by x and completing the square.
400 Easy
400 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 and comparing coefficients, the value of k in the given integral is found to be 2.
401 Easy
401 of 995
Application +4 | -1
∫cos⁡{2tan⁡−11−x1+x}dx\int \cos \left\{2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}\right\} d x is equal to
View answer and explanation
Correct option: B
12x2+k\frac{1}{2} x^2+k
Option B (12x2+k\frac{1}{2} x^2+k) is correct because it is the result of simplifying the integral after trigonometric substitution.
402 Easy
402 of 995
Application +4 | -1
∫1sin⁡(x−a)sin⁡(x−b)dx=\int \frac{1}{\sin (x-a) \sin (x-b)} d x=
View answer and explanation
Correct option: A
1sin⁡(a−b)log⁡∣sin⁡(x−a)sin⁡(x−b)∣+C\frac{1}{\sin (a-b)} \log \left|\frac{\sin (x-a)}{\sin (x-b)}\right|+C
Option A (1sin⁡(a−b)log⁡∣sin⁡(x−a)sin⁡(x−b)∣+C\frac{1}{\sin (a-b)} \log \left|\frac{\sin (x-a)}{\sin (x-b)}\right|+C) is correct because by using the sine subtraction formula and simplifying, the integral evaluates to this expression.
403 Easy
403 of 995
Application +4 | -1
If f(x)=cos⁡xf(x)=\cos x then ∫(2(f(x))2−1)(4(f(x))3−3\int\left(2(f(x))^2-1\right)\left(4(f(x))^3-3\right. f(x))dxf(x)) \mathrm{d} x is equal to
View answer and explanation
Correct option: A
110sin⁡5x+12sin⁡x+C\frac{1}{10} \sin 5 x+\frac{1}{2} \sin x+C
The correct answer is A because using trigonometric identities to simplify the integrand and then integrating yields the result.
404 Medium
404 of 995
Analyzing +4 | -1
If ∫  2x2+  3(x2−1)  (x2+4)  dx=a  log⁡  (x−1x+1)  +  b  tan⁡−1  (x2)+c\int \,\,\frac{{2{x^2} + \,\,3}}{{\left( {{x^2} - 1} \right)\,\,\left( {{x^2} + 4} \right)}}\,\,dx = a\,\,\log \,\,\left( {\frac{{x - 1}}{{x + 1}}} \right)\,\, + \,\,b\,\,{\tan ^{ - 1}}\,\,\left( {\frac{x}{2}} \right) + c, then the value of aa and bb are
View answer and explanation
Correct option: D
(12,  12)\left( {\frac{1}{2},\,\,\frac{1}{2}} \right)
Option D (12,  12)\left( {\frac{{1}}{{2}},\,\,\frac{{1}}{{2}}} \right) is correct because by partial fraction decomposition and integration, we find that a=12a = \frac{1}{2} and b=12b = \frac{1}{2}.
405 Medium
405 of 995
Analyzing +4 | -1
∫(esin⁡xesin⁡x+x)2(1x−cos⁡x)dx=log⁡e(f(x))+g(x)+c\int\left(\frac{e^{\sin x}}{e^{\sin x}+x}\right)^2\left(\frac{1}{x}-\cos x\right) d x=\log _e(f(x))+g(x)+c; where cc is constant, then f(π)−g(π)f(\pi)-g(\pi) is
View answer and explanation
Correct option: B
π−1π+1\frac{\pi-1}{\pi+1}
Option B is correct because after the substitution and integration, the expression simplifies to the given form, and evaluating f(π) - g(π) yields (π-1)/(π+1).

Explore Other Chapters in MATHEMATICS

View All →
SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs DIFFERENTIAL EQUATIONS 977 MCQs
406 Medium
406 of 995
Application +4 | -1
∫cos⁡3xelog⁡(sin⁡x)dx\int \cos ^3 x e^{\log (\sin x)} d x is equal to
View answer and explanation
Correct option: B
−cos⁡4x4+c-\frac{\cos ^4 x}{4}+c
Option B is correct because elog⁡(sin⁡x)=sin⁡xe^{\log(\sin x)} = \sin x, so the integral becomes ∫cos⁡3xsin⁡xdx\int \cos^3 x \sin x dx, which is solved by substituting t=cos⁡xt = \cos x.
407 Easy
407 of 995
Analyzing +4 | -1
Let f:R→Rf:R \to R be a differentiable function such that f(π4)=2,  f(π2)=0f\left( {\frac{\pi }{4}} \right) = \sqrt 2 ,\,\,f\left( {\frac{\pi }{2}} \right) = 0 and f′(π2)=1f'\left( {\frac{\pi }{2}} \right) = 1 and let g(x)=∫xπ4(f′(t)sec⁡t+tan⁡tsec⁡t(t))dtg\left( x \right) = \int_x^{\frac{\pi }{4}} {\left( {f'\left( t \right)\sec t + \tan t\sec t\left( t \right)} \right)} dt for x∈[π4,π2).x \in \left[ {\frac{\pi }{4},\frac{\pi }{2}} \right). Then lim⁡x→(π2)−g(x)\lim\limits_{x \to {{\left( {\frac{\pi }{2}} \right)}^ - }} g\left( x \right) is equal to
View answer and explanation
Correct option: B
3
Option B (3) is correct because the limit of g(x)g(x) as xx approaches π2\frac{\pi}{2} from the left is found by substituting x=π2−hx = \frac{\pi}{2} - h and applying L'Hôpital's rule, resulting in 3.
408 Easy
408 of 995
Application +4 | -1
∫sin⁡−1(3x−4x3)dx=\int \sin ^{-1}\left(3 x-4 x^3\right) d x=
View answer and explanation
Correct option: D
3[xsin⁡−1x+1−x2]+c3\left[x \sin ^{-1} x+\sqrt{1-x^2}\right]+c
Option D (3[xsin⁡−1x+1−x2]+c3\left[x \sin ^{-1} x+\sqrt{1-x^2}\right]+c) is correct because substituting x=sin⁡θx = \sin \theta simplifies the integral, and integration by parts yields the correct result.
409 Easy
409 of 995
Application +4 | -1
The value of ∫sin⁡nxcos⁡n+6x3dx\int \sqrt[3]{\frac{\sin ^n x}{\cos ^{n+6} x}} d x, is
View answer and explanation
Correct option: B
3n+3tan⁡n/3+1x+C\frac{3}{n+3} \tan ^{n / 3+1} x+C
Option B is correct because the integral simplifies to (3/(n+3))tan(n/3 + 1)(x) + C after recognizing the derivative of tan(x).
410 Easy
410 of 995
Application +4 | -1
∫[sin⁡(log⁡x)+cos⁡(log⁡x)]dx=\int[\sin (\log x)+\cos (\log x)] d x=
View answer and explanation
Correct option: D
xsin⁡(log⁡x)+cx \sin (\log x)+c
Option D (xsin⁡(log⁡x)+cx \sin (\log x)+c) is correct because it is the result of integrating the given expression using integration by parts.
411 Medium
411 of 995
Analyzing +4 | -1
The value of ∫(x−1)(x+1)x3+x2+xdx\int \frac{(x-1)}{(x+1) \sqrt{x^3+x^2+x}} d x, is
View answer and explanation
Correct option: C
2tan⁡−1x2+x+1x+C2 \tan ^{-1} \sqrt{\frac{x^2+x+1}{x}}+C
Option C (2tan⁡−1x2+x+1x+C2 \tan ^{-1} \sqrt{\frac{x^2+x+1}{x}}+C) is correct because the integral simplifies to this form after appropriate substitutions and trigonometric manipulations.
412 Easy
412 of 995
Application +4 | -1
∫1x1−x3dx\int \frac{1}{x \sqrt{1-x^3}} d x is equal to
View answer and explanation
Correct option: A
13log⁡∣1−x3−11−x3+1∣+C\frac{1}{3} \log \left|\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}\right|+C
Option A (13log⁡∣1−x3−11−x3+1∣+C\frac{{1}}{{3}} \log \left| \frac{{\sqrt{{1-x^3}}-1}}{{\sqrt{{1-x^3}}+1}} \right|+C) is correct because it is the result of the integration after appropriate substitution and simplification.
413 Easy
413 of 995
Application +4 | -1
Let f  (x)=∫  1(1+x2)3/2  dxf\,\,\left( x \right) = \int \,\,\frac{1}{{{{\left( {1 + {x^2}} \right)}^{3/2}}}}\,\,dx and f  (0)=0f\,\,\left( 0 \right) = 0, then f  (1)=f\,\,\left( 1 \right) =
View answer and explanation
Correct option: B
12\frac{1}{{\sqrt 2 }}
Option B (12\frac{1}{{\sqrt 2 }}) is correct because the integral of the given function is evaluated, and the constant of integration is determined using the initial condition f(0) = 0.
414 Easy
414 of 995
Application +4 | -1
∫x2(3)x3+1dx=\int x^2(3)^{x^3+1} d x=
View answer and explanation
Correct option: B
(3)x3log⁡3+c\frac{(3)^{x^3}}{\log 3}+c
Option B is correct because the integral can be solved using u-substitution and the integral of an exponential function, resulting in (3)x3log⁡3+c\frac{(3)^{x^3}}{\log 3}+c.
415 Medium
415 of 995
Application +4 | -1
If ∫dx1+sin⁡x=tan⁡(x2+a)+c\int \frac{d x}{1+\sin x}=\tan \left(\frac{x}{2}+a\right)+c, then aa is equal to
View answer and explanation
Correct option: D
−π4-\frac{\pi}{4}
Option D (−π4-\frac{\pi}{4}) is correct because the integral evaluates to tan⁡(x2−π4)+c\tan(\frac{x}{2} - \frac{\pi}{4}) + c, thus a=−π4a = -\frac{\pi}{4}.
416 Easy
416 of 995
Analyzing +4 | -1
The integral ∫1(1+x)x−x2dx\int \frac{1}{(1+\sqrt{x}) \sqrt{x-x^2}} d x is equal to (where CC is the constant of integration)
View answer and explanation
Correct option: B
−21−x1+x-2 \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}
Option B is correct because after using the trigonometric substitution x=cos⁡2θx = \cos^2 \theta, the integral simplifies and evaluates to −21−x1+x+C-2 \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}+C.
417 Easy
417 of 995
Analyzing +4 | -1
∫  x2+  1  [log⁡  (x2+1)−2  log⁡  x]x4  dx\int \,\,\frac{{\sqrt {{x^2} + \,\,1} \,\,\left[ {\log \,\,\left( {{x^2} + 1} \right) - 2\,\,\log \,\,x} \right]}}{{{x^4}}}\,\,dx is equal to
View answer and explanation
Correct option: B
−13  (1  +  1x2)3/2  [log⁡  (1  +  1x2)  −  23]+C- \frac{1}{3}\,\,{\left( {1\,\, + \,\,\frac{1}{{{x^2}}}} \right)^{3/2}}\,\,\left[ {\log \,\,\left( {1\,\, + \,\,\frac{1}{{{x^2}}}} \right)\,\, - \,\,\frac{2}{3}} \right] + C
Option B (−13 (1 + 1x2)3/2  [log⁡  (1 + 1x2) −  23]+C- \frac{1}{3}\,{\left( {1\, + \,\frac{1}{{{{x^2}}}}} \right)^{3/2}}\,\,\left[ {\log \,\,\left( {1\, + \,\frac{1}{{{{x^2}}}}} \right)\, - \,\,\frac{2}{3}} \right] + C) is correct because after substitution and integration by parts, the integral evaluates to the given expression.
418 Medium
418 of 995
Analyzing +4 | -1
∫−11  17 x5  −  x4  +  29 x3−  31 x  +  1x2+  1  dx    is\int \nolimits_{ - 1}^1 \,\,\frac{{17\,{x^5}\,\, - \,\,{x^4}\,\, + \,\,29\,{x^3} - \,\,31\,x\,\, + \,\,1}}{{{x^2} + \,\,1}}\,\,dx\,\,\;\mathrm{is}
View answer and explanation
Correct option: C
43\frac{4}{3}
Option C (43\frac{4}{3}) is correct because the integral of the odd part is zero, and the integral of the even part is calculated to be 43\frac{4}{3}.
419 Easy
419 of 995
Application +4 | -1
∫xlog⁡xdx=\int x \log x d x=
View answer and explanation
Correct option: B
x22log⁡x−x24+c\frac{x^2}{2} \log x-\frac{x^2}{4}+c
Option B (x22log⁡x−x24+c\frac{x^2}{2} \log x-\frac{x^2}{4}+c) is correct because it is the result of integrating xlog⁡xx \log x using integration by parts.
420 Easy
420 of 995
Analyzing +4 | -1
If ∫sin⁡−1(x1+x)dx = A(x) tan⁡−1(x) + B (x) + C, where C\int {{{\sin }^{ - 1}}\left( {\sqrt {\frac{x}{{1 + x}}} } \right)dx\, = \,A(x)\,{{\tan }^{ - 1}}\left( {\sqrt x } \right)\, + \,B\,(x)\, + \,C,\,\mathrm{where}\,C} is a constant of integration, then the ordered pair (A(x), B (x))\left( {A(x),\,B\,(x)} \right) can be
View answer and explanation
Correct option: C
(x+1, −x)\left( {x + 1,\, - \sqrt x } \right)
Option C ((x+1, −x)\left( {x + 1,\, - \sqrt x } \right)) is correct because after evaluating the integral using trigonometric substitution and integration by parts, the result matches the given form with A(x) = x+1 and B(x) = -x\sqrt{x}.
Revision notes and key takeaways

Review INDEFINITE INTEGRATION in your textbook or class notes, then use this set to check your understanding.

Showing 391-420 of 995 questions

Continue Your Practice

📚 More Sets in This Chapter

Try different difficulty levels to master this topic

→ ENGINEERING MATHEMATICS - INDEFINITE INTEGRATION MCQs (Easy) 746 MCQs → ENGINEERING MATHEMATICS - INDEFINITE INTEGRATION MCQs (Hard) 15 MCQs → ENGINEERING MATHEMATICS - INDEFINITE INTEGRATION MCQs (Medium) 411 MCQs
🎯 Next Chapters to Practice

Continue building your subject mastery

→ SEQUENCE OF SERIES 1764 MCQs → DEFINITE INTEGRATION 1443 MCQs → MATRICES 1363 MCQs → PROBABILITY 1146 MCQs

Your practice

0 / 995 answered · 995 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Practice settings

Type at least 2 characters or jump directly:
🩺 NEET MCQs 🔬 JEE MCQs 🎯 JEE College Predictor Study guides 📊 Dashboard
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.

Continue with Free Account →