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JEE • MATHEMATICS • INDEFINITE INTEGRATION

Indefinite integration practice · Easy

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181 Easy
181 of 746
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.
182 Easy
182 of 746
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.
183 Easy
183 of 746
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.
184 Easy
184 of 746
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.
185 Easy
185 of 746
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.
186 Easy
186 of 746
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).
187 Easy
187 of 746
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.
188 Easy
188 of 746
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.
189 Easy
189 of 746
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.
190 Easy
190 of 746
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.
191 Easy
191 of 746
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.
192 Easy
192 of 746
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.
193 Easy
193 of 746
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.
194 Easy
194 of 746
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.
195 Easy
195 of 746
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.

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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
196 Easy
196 of 746
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.
197 Easy
197 of 746
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.
198 Easy
198 of 746
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.
199 Easy
199 of 746
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.
200 Easy
200 of 746
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)).
201 Easy
201 of 746
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.
202 Easy
202 of 746
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.
203 Easy
203 of 746
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.
204 Easy
204 of 746
Application +4 | -1
∫x2−1x32x4−2x2+1dx\int \frac{x^2-1}{x^3 \sqrt{2 x^4-2 x^2+1}} d x is equal to
View answer and explanation
Correct option: D
2x4−2x2+12x2+C\frac{\sqrt{2 x^4-2 x^2+1}}{2 x^2}+C
Option D is correct because by substituting z=2−2x2+1x4z = 2 - \frac{2}{x^2} + \frac{1}{x^4}, the integral simplifies to 2x4−2x2+12x2+C\frac{\sqrt{2 x^4-2 x^2+1}}{2 x^2}+C.
205 Easy
205 of 746
Application +4 | -1
If I=∫sec⁡−1xdxI=\int \sec ^{-1} \sqrt{x} d x, then II equals
View answer and explanation
Correct option: C
xsec⁡−1x−x−1+Cx \sec ^{-1} \sqrt{x}-\sqrt{x-1}+C
Option C (\xsec⁡−1x−x−1+Cx \sec^{-1} \sqrt{x} - \sqrt{x-1} + C is correct. The integral is solved by using substitution and integration by parts.
206 Easy
206 of 746
Analyzing +4 | -1
Let f(x)={sin⁡(tan⁡−1x)+sin⁡(cot⁡−1x)}2−1f(x)=\left\{\sin \left(\tan ^{-1} x\right)+\sin \left(\cot ^{-1} x\right)\right\}^2-1, where ∣x∣>1|x|>1 and dydx=12ddx(sin⁡−1f(x))\frac{d y}{d x}=\frac{1}{2} \frac{d}{d x}\left(\sin ^{-1} f(x)\right). If y(3)=π3y(\sqrt{3})=\frac{\pi}{3}, then y(−3)=y(-\sqrt{3})=
View answer and explanation
Correct option: B
−π3-\frac{\pi}{3}
Option B (-π3\frac{\pi}{3}) is correct because after simplifying f(x) and using the given condition, we find y(-3\sqrt{3}) = -π3\frac{\pi}{3}.
207 Easy
207 of 746
Application +4 | -1
∫  cos⁡−1  (1x)  dx\int \,\,{\cos ^{ - 1}}\,\,\left( {\frac{1}{x}} \right)\,\,dx equals
View answer and explanation
Correct option: B
x  sec⁡−1  x  −  cosh⁡−1   x  +  cx\,\,{\sec ^{ - 1}}\,\,x\,\, - \,\,{\cosh ^{ - 1}}\,\,\,x\,\, + \,\,c
Option B (x sec⁡−1 x − cosh⁡−1 x + cx\,{\sec ^{ - 1}}\,x\, - \,{\cosh ^{ - 1}}\,x\, + \,c) is correct because it is the result of integration by parts and simplification using inverse hyperbolic functions.
208 Easy
208 of 746
Application +4 | -1
If ∫(sin⁡2x−cos⁡2x)dx=12sin⁡(2x+k)+C\int(\sin 2 x-\cos 2 x) \mathrm{d} x=\frac{1}{\sqrt{2}} \sin (2 x+k)+C then
View answer and explanation
Correct option: B
k=5π/4k=5 \pi / 4
Option B is correct because after integrating by parts and simplifying, the integral evaluates to (x−1)log⁡(1−x)−12x−x+C(x-1) \log (1-\sqrt{x})-\frac{1}{2} x-\sqrt{x}+C.
209 Easy
209 of 746
Analyzing +4 | -1
Let ff be a non-negative function defined on the interval [0,1]\left[ {0,1} \right]. If ∫0x  1  −  ( f′(t))2  dt=∫0x  f  (t)  dt,  0≤  x  ≤1\int \nolimits_0^x \,\,\sqrt {1\,\, - \,\,{{\left( {\,f'\left( t \right)} \right)}^2}} \,\,dt = \int \nolimits_0^x \,\,f\,\,\left( t \right)\,\,dt,\,\,0 \le \,\,x\,\, \le 1 and f  (0)=0f\,\,\left( 0 \right) = 0, then

View answer and explanation
Correct option: C
f  (12)  <  12f\,\,\left( {\frac{1}{2}} \right)\,\, < \,\,\frac{1}{2} and f  (13)  <  13f\,\,\left( {\frac{1}{3}} \right)\,\, < \,\,\frac{1}{3}
Option C (f  (12)  <  12f\,\,\left( {\frac{1}{2}} \right)\,\,< \,\,\frac{1}{2} and f  (13)  <  13f\,\,\left( {\frac{1}{3}} \right)\,\,< \,\,\frac{1}{3}) is correct because f(x)=sin⁡(x)f(x) = \sin(x) and sin⁡(x)<x\sin(x) < x for all positive x.
210 Easy
210 of 746
Application +4 | -1
∫  x4  +  x2+  1x2  −  x  +  1  dx\int \,\,\frac{{{x^4}\,\, + \,\,{x^2} + \,\,1}}{{{x^2}\,\, - \,\,x\,\, + \,\,1}}\,\,dx is equal to
View answer and explanation
Correct option: B
x33  +  x22  +  x  +  c\frac{{{x^3}}}{3}\,\, + \,\,\frac{{{x^2}}}{2}\,\, + \,\,x\,\, + \,\,c
Option B (x33 +  x22 + x + c\frac{{{x^3}}}{3}\, + \,\,\frac{{{x^2}}}{2}\, + \,x\, + \,c) is correct because the integral is solved by polynomial long division and then integrating the resulting polynomial.
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