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JEE • MATHEMATICS • INDEFINITE INTEGRATION

Indefinite integration practice · Mixed difficulty

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241 Easy
241 of 995
Application +4 | -1
The value of ∫1a  [x]  f′(x) dx,  a>1\int \nolimits_1^a \,\,\left[ x \right]\,\,f'\left( x \right)\,dx,\,\,a > 1 where [x]\left[ x \right] denotes the greatest integer not exceedingx, is
View answer and explanation
Correct option: A
[a]  f(a)  −  {  f(1)  +  f  (2)  +  ...  +f  ([a])}\left[ a \right]\,\,f\left( a \right)\,\, - \,\,\left\{ {\,\,f\left( 1 \right)\,\, + \,\,f\,\,\left( 2 \right)\,\, + \,\,...\,\, + f\,\,\left( {\left[ a \right]} \right)} \right\}
Option A is correct because the integral is evaluated by splitting the range based on the greatest integer function and then summing the results.
242 Easy
242 of 995
Application +4 | -1
∫(x+1)  (x+2)7 (x+3) dx\int \left( {x + 1} \right)\,\,{\left( {x + 2} \right)^7}\,\left( {x + 3} \right)\,dx is equal to
View answer and explanation
Correct option: A
(x+2)1010  −  (x+2)88  +  c\frac{{{{\left( {x + 2} \right)}^{10}}}}{{10}}\,\, - \,\,\frac{{{{\left( {x + 2} \right)}^8}}}{8}\,\, + \,\,c
The correct answer is ((x + 2)10)/10 - ((x + 2)8)/8 + c because by rewriting the expression and integrating term by term, we arrive at this result.
243 Easy
243 of 995
Application +4 | -1
∫log⁡(x+x2−1)dx=\int \log \left(x+\sqrt{x^2-1}\right) d x=
View answer and explanation
Correct option: A
xlog⁡e(x+x2−1)−x2−1+cx \log _e\left(x+\sqrt{x^2-1}\right)-\sqrt{x^2-1}+c
Option A (xlog⁡e(x+x2−1)−x2−1+cx \log _e(x+\sqrt{x^2-1})-\sqrt{x^2-1}+c) is correct because it is the result of integration by parts applied to the given integral.
244 Easy
244 of 995
Application +4 | -1
∫  log⁡e  (tan⁡  x)sin⁡  x  cos⁡  x  dx\int \,\,\frac{{{{\log }_e}\,\,\left( {\tan \,\,x} \right)}}{{\sin \,\,x\,\,\cos \,\,x}}\,\,dx is equal to
View answer and explanation
Correct option: B
 12  (log⁡e tan⁡  x)2+  c\,\frac{1}{2}\,\,{\left( {{{\log }_e}\,\tan \,\,x} \right)^2} + \,\,c
Option B is correct because the integral simplifies to \frac{1}{2}\,{\left( {\log _e}\,\tan \,x} \right)^2} + \,c\frac{1}{2}\,{\left( {\log _e}\,\tan \,x} \right)^2} + \,c after applying u-substitution twice.
245 Medium
245 of 995
Application +4 | -1
∫sin⁡x+cos⁡x9+16sin⁡2xdx\int \frac{\sin x+\cos x}{9+16 \sin 2 x} d x is equal to
View answer and explanation
Correct option: A
140ln⁡∣5+4(sin⁡x−cos⁡x)5−4(sin⁡x−cos⁡x)∣+c\frac{1}{40} \ln \left|\frac{5+4(\sin x-\cos x)}{5-4(\sin x-\cos x)}\right|+c
Option A (140ln⁡∣5+4(sin⁡x−cos⁡x)5−4(sin⁡x−cos⁡x)∣+c\frac{{1}}{{40}} \ln \left| {\frac{{5 + 4(\sin x - \cos x)}}{{5 - 4(\sin x - \cos x)}}} \right| + c) is correct because the integral is solved by substitution using t=sin⁡x−cos⁡xt = \sin x - \cos x.
246 Easy
246 of 995
Application +4 | -1
∫  x49   tan⁡−1  (x50)(1+x100)  dx=k [tan⁡−1  (x50)]2+c\int \,\,\frac{{{x^{49}}\,\,\,{{\tan }^{ - 1}}\,\,\left( {{x^{50}}} \right)}}{{\left( {1 + {x^{100}}} \right)}}\,\,dx = k\,{\left[ {{{\tan }^{ - 1}}\,\,\left( {{x^{50}}} \right)} \right]^2} + c, then kk is equal to
View answer and explanation
Correct option: C
1100\frac{1}{{100}}
Option C (1100\frac{1}{{100}}) is correct because it is the coefficient of the squared arctangent term after performing the integration by substitution.
247 Easy
247 of 995
Application +4 | -1
∫1xsec⁡2(log⁡x)dx=\int \frac{1}{x} \sec ^2(\log x) d x=
View answer and explanation
Correct option: A
tan⁡(log⁡x)+c\tan (\log x)+c
Option A (tan⁡(log⁡x)+c\tan (\log x)+c) is correct because it is the result of a u-substitution and applying the standard integral of sec⁡2(t)\sec^2(t).
248 Easy
248 of 995
Understanding +4 | -1
Assertion : The area bounded by the curve y=cos⁡xy=\cos x in I quadrant with the coordinate axes is 1 sq. unit.
Reason : ∫0π/2cos⁡xdx=1\int\limits_{0}^{\pi / 2} \cos x d x=1
View answer and explanation
Correct option: A
Assertion is correct, Reason is correct; Reason is a correct explanation for assertion.
Assertion is correct, Reason is correct; Reason is a correct explanation for assertion because the area under the curve y=cos(x) from 0 to pi/2 is indeed 1, which is calculated by the definite integral.
249 Easy
249 of 995
Application +4 | -1
The value of ∫(tan⁡−1(sin⁡x+1))cos⁡x(3+2sin⁡x−cos⁡2x)dx\int \frac{\left(\tan ^{-1}(\sin x+1)\right) \cos x}{\left(3+2 \sin x-\cos ^2 x\right)} d x is (where cc is the constant of integration)
View answer and explanation
Correct option: C
(tan⁡−1(sin⁡x+1))22+c\frac{\left(\tan ^{-1}(\sin x+1)\right)^2}{2}+c
Option C, (tan⁡−1(sin⁡x+1))22+c\frac{\left(\tan ^{-1}(\sin x+1)\right)^2}{2}+c, is correct because it results from the substitution and integration of the given expression.
250 Easy
250 of 995
Application +4 | -1
∫x5⋅ex2dx=\int x^5 \cdot e^{x^2} d x=
View answer and explanation
Correct option: A
12x4ex2−x2ex2+ex2+c\frac{1}{2} x^4 e^{x^2}-x^2 e^{x^2}+e^{x^2}+c
Option A (12x4ex2−x2ex2+ex2+c\frac{1}{2} x^4 e^{x^2}-x^2 e^{x^2}+e^{x^2}+c) is correct because substituting x2=tx^2 = t simplifies the integral, and integration by parts yields the correct result.
251 Easy
251 of 995
Application +4 | -1
∫x2+x−1x2+x−6dx=\int \frac{x^2+x-1}{x^2+x-6} d x=
View answer and explanation
Correct option: B
x−log⁡(x+3)+log⁡(x−2)+cx-\log (x+3)+\log (x-2)+c
Option B, x−log⁡(x+3)+log⁡(x−2)+cx-\log (x+3)+\log (x-2)+c, is the correct answer because it results from polynomial long division, partial fraction decomposition, and integration of the given expression.
252 Medium
252 of 995
Application +4 | -1
∫e2x(−sin⁡x+2cos⁡x)dx=\int e^{2 x}(-\sin x+2 \cos x) d x=
View answer and explanation
Correct option: D
e2xcos⁡x+ce^{2 x} \cos x+c
Option D is correct because the integral of e(2x)(-sin x + 2cos x) dx simplifies to e(2x)cos x + c using the property of integrals of the form ∫e(kx){kf(x)+f'(x)}dx = e(kx)f(x) + c.
253 Easy
253 of 995
Application +4 | -1
The integral ∫2x12+5x9(x5+x3+1)3dx\int \frac{2 x^{12}+5 x^9}{\left(x^5+x^3+1\right)^3} d x is equal to (where CC is a constant of integration)
View answer and explanation
Correct option: B
x102(x5+x3+1)2+C\frac{x^{10}}{2\left(x^5+x^3+1\right)^2}+C
Option B is correct because the integral simplifies to x102(x5+x3+1)2+C\frac{x^{10}}{2\left(x^5+x^3+1\right)^2}+C after dividing by x15x^{15} and using substitution.
254 Medium
254 of 995
Understanding +4 | -1
If ddx  f(x)=g(x)\frac{d}{{dx}}\,\,f\left( x \right) = g\left( x \right) for a≤x≤ba \le x \le b then, ∫ab f(x)  g  (x)  dx\int \limits_a^b \,f\left( x \right)\,\,g\,\,\left( x \right)\,\,dx equals
View answer and explanation
Correct option: C
[ f(b) ]2 −  [ f(a) ]22\frac{{{{\left[ {\,f\left( b \right)\,} \right]}^2}\, - \,\,{{\left[ {\,f\left( a \right)\,} \right]}^2}}}{2}
Option C, \frac{{{{\left[ {\,f\left( b \right)\} \right]}^2}\, - \,{{\left[ {\,f\left( a \right)\} \right]}^2}}}{2}\frac{{{{\left[ {\,f\left( b \right)\} \right]}^2}\, - \,{{\left[ {\,f\left( a \right)\} \right]}^2}}}{2}, is correct because the integral can be solved by recognizing that g(x)dx=df(x)g(x)dx = df(x).
255 Easy
255 of 995
Application +4 | -1
∫dxx(x4+1)\int \frac{\mathrm{d} x}{x\left(x^4+1\right)} is equal to
View answer and explanation
Correct option: B
14log⁡(x4x4+1)+C\frac{1}{4} \log \left(\frac{x^4}{x^4+1}\right)+C
Option B (14log⁡(x4x4+1)+C\frac{1}{4} \log \left(\frac{x^4}{x^4+1}\right)+C) is correct because it is the result of integrating the given expression using substitution and partial fractions.

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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
256 Easy
256 of 995
Application +4 | -1
∫cos⁡x(1+sin⁡x)(2+sin⁡x)dx=\int \frac{\cos x}{(1+\sin x)(2+\sin x)} d x=
View answer and explanation
Correct option: C
log⁡1+sin⁡x2+sin⁡x+c\log \frac{1+\sin x}{2+\sin x}+c
Option C is correct because the integral is solved by substituting sin⁡x=t\sin x = t and using partial fraction decomposition.
257 Medium
257 of 995
Application +4 | -1
∫dx(x−3)2x2−6x+4\int \frac{d x}{(x-3)^2 \sqrt{x^2-6 x+4}} is equal to
View answer and explanation
Correct option: A
x2−6x+45∣x−3∣+c\frac{\sqrt{x^2-6 x+4}}{5|x-3|}+c
Option A (x2−6x+45∣x−3∣+c\frac{\sqrt{x^2-6 x+4}}{5|x-3|}+c) is correct because the integral, after substitution and simplification, results in this expression.
258 Medium
258 of 995
Application +4 | -1
The value of ∫1sin⁡(x−π3)cos⁡xdx\int \frac{1}{\sin \left(x-\frac{\pi}{3}\right) \cos x} d x, is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the integral evaluates to 2log⁡∣sin⁡(x−π3)sec⁡x∣+C2 \log |\sin (x-\frac{\pi}{3}) \sec x|+C, which is not among the given options.
259 Easy
259 of 995
Application +4 | -1
∫  1(1+x2)  1−x2  dx\int \,\,\frac{1}{{\left( {1 + {x^2}} \right)\,\,\sqrt {1 - {x^2}} }}\,\,dx is equal to
View answer and explanation
Correct option: C
12  tan⁡−1(2x1  −  x2)\frac{1}{{\sqrt 2 }}\,\,{\tan ^{ - 1}}\left( {\frac{{\sqrt 2 x}}{{\sqrt {1\,\, - \,\,{x^2}} }}} \right)
Option C is correct because the integral is solved using a series of substitutions, resulting in the given expression.
260 Easy
260 of 995
Application +4 | -1
∫ex−  1 dx\int \sqrt {{e^x} - \,\,1} \,dx is equal to
View answer and explanation
Correct option: A
2  [ex−1−tan⁡−1  ex−1]+c2\,\,\left[ {\sqrt {{e^x} - 1} - {{\tan }^{ - 1}}\,\,\sqrt {{e^x} - 1} } \right] + c
Option A is correct because the integral of ex−1\sqrt{{e^x} - 1} is found using substitution and trigonometric identities, resulting in 2[ex−1−tan⁡−1 ex−1]+c2[\sqrt{{e^x} - 1} - {\tan}^{ - 1}\,\sqrt{{e^x} - 1}] + c.
261 Easy
261 of 995
Application +4 | -1
∫1x(x7+1)dx\int \frac{1}{x\left(x^7+1\right)} d x is equal to
View answer and explanation
Correct option: B
17log⁡(x7x7+1)+C\frac{1}{7} \log \left(\frac{x^7}{x^7+1}\right)+C
Option B, 17log⁡(x7x7+1)+C\frac{1}{7} \log \left(\frac{x^7}{x^7+1}\right)+C, is correct because it results from a u-substitution and partial fraction decomposition of the integral.
262 Medium
262 of 995
Application +4 | -1
∫  x2  +  a2   dx\int \,\,\sqrt {{x^2}\,\, + \,\,{a^2}} \;\,dx equals
View answer and explanation
Correct option: B
x2  x2+  a2  +  a22  log⁡  {x+x2+  a2}  +  c\frac{x}{2}\,\,\sqrt {{x^2} + \,\,{a^2}} \,\, + \,\,\frac{{{a^2}}}{2}\,\,\log \,\,\left\{ {x + \sqrt {{x^2} + \,\,{a^2}} } \right\}\,\, + \,\,c
Option B (x2 x2+ a2 + a22 log⁡ {x+x2+ a2} + c\frac{x}{2}\,\sqrt {{x^2} + \,{a^2}} \, + \,\frac{{{a^2}}}{2}\,\log \,\left\{ {x + \sqrt {{x^2} + \,{a^2}} } \right\}\, + \,c) is correct because it is the standard integral formula for ∫ x2 + a2   dx\int \,\sqrt {{x^2}\, + \,{a^2}} \;\,dx, derived using integration by parts.
263 Easy
263 of 995
Application +4 | -1
If ∫  dx(x  +  100)  x  +  99=f  (x)  +  c\int \,\,\frac{{dx}}{{\left( {x\,\, + \,\,100} \right)\,\,\sqrt {x\,\, + \,\,99} }} = f\,\,\left( x \right)\,\, + \,\,c, then f  (x)f\,\,\left( x \right) is
View answer and explanation
Correct option: C
2  tan⁡−1  (x+99)2\,\,{\tan ^{ - 1}}\,\,\left( {\sqrt {x + 99} } \right)
Option C is correct because after performing the substitution, the integral is simplified to a standard arctangent form, leading to the given result.
264 Medium
264 of 995
Analyzing +4 | -1
∫dx(x2+x+1)x2+x−1=\int \frac{d x}{\left(x^2+x+1\right) \sqrt{x^2+x-1}}=
View answer and explanation
Correct option: A
4∫du8−3u24 \int \frac{d u}{8-3 u^2} where u=1−z2,z=1t and x+12=t52u=\sqrt{1-z^2}, z=\frac{1}{t} \text{ and } x+\frac{1}{2}=\frac{t \sqrt{5}}{2}
Option A (4∫du8−3u24 \int \frac{{d u}}{{8-3 u^2}} where u=1−z2,z=1tu=\sqrt{{1-z^2}}, z=\frac{{1}}{{t}} and x+12=t52x+\frac{{1}}{{2}}=\frac{{t \sqrt{{5}}}}{{2}}) is correct because the given substitutions lead to the correct simplified integral.
265 Easy
265 of 995
Application +4 | -1
∫etan⁡−1x(1+x+x21+x2)dx\int e^{\tan ^{-1} x}\left(\frac{1+x+x^2}{1+x^2}\right) d x is equal to
View answer and explanation
Correct option: A
xetan⁡−1x+cx e^{\tan ^{-1} x}+c
Option A (xetan⁡−1x+cx e^{\tan ^{-1} x}+c) is correct because 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 allows for direct integration after substitution.
266 Easy
266 of 995
Application +4 | -1
∫xcos⁡xdx=\int x \cos x d x=
View answer and explanation
Correct option: A
xsin⁡x+cos⁡xx \sin x+\cos x
Option A (xsin⁡x+cos⁡xx \sin x+\cos x) is correct because it is the result of integrating xcos⁡xx \cos x using integration by parts.
267 Easy
267 of 995
Application +4 | -1
If ∫ex(1−sin⁡x1−cos⁡x)dx=f(x)+\int e^x\left(\frac{1-\sin x}{1-\cos x}\right) d x=f(x)+ Constant, then f(x)f(x) is equal to
View answer and explanation
Correct option: C
−excot⁡(x2)+C-e^x \cot \left(\frac{x}{2}\right)+C
Option C (−excot⁡(x2)+C-e^x \cot \left(\frac{x}{2}\right)+C) is correct because the integral simplifies to this form after using trigonometric identities and recognizing the pattern of ex(f(x)+f′(x))e^x(f(x) + f'(x)).
268 Easy
268 of 995
Application +4 | -1
∫  1sin⁡2  x  .  cos⁡2  x    dx\int \,\,\frac{1}{{{{\sin }^2}\,\,x\,\,.\,\,{{\cos }^2}\,\,x}}\;\,\,dx is equal to
View answer and explanation
Correct option: D
tan⁡  x  −  cot⁡  x  +  c\tan \,\,x\,\, - \,\,\cot \,\,x\,\, + \,\,c
Option D (tan⁡ x − cot⁡ x + c\tan \,x\, - \,\cot \,x\, + \,c) is correct because the integral simplifies to the sum of the integrals of sec²x and csc²x, which are tan x and -cot x, respectively.
269 Easy
269 of 995
Application +4 | -1
∫tan⁡4x dx=Atan⁡3x+Btan⁡x+f(x)\int \tan ^4 x \,d x=A \tan ^3 x+B \tan x+f(x), then
View answer and explanation
Correct option: A
A=13,B=−1,f(x)=x+CA=\frac{1}{3}, B=-1, f(x)=x+C
Option A is correct because by integrating tan⁡4x\tan^4 x, we find that A=13,B=−1,f(x)=x+CA=\frac{1}{3}, B=-1, f(x)=x+C.
270 Easy
270 of 995
Application +4 | -1
∫0π/2cos⁡xesin⁡xdx\int\limits_{0}^{\pi / 2} \cos x e^{\sin x} d x is equal to
View answer and explanation
Correct option: B
e−1e-1
Option B (e−1e-1) is correct because the integral can be solved using substitution, resulting in esin⁡xe^{\sin x} evaluated from 0 to π/2, which gives e - 1.
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