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JEE • MATHEMATICS • INDEFINITE INTEGRATION

Indefinite integration practice · Mixed difficulty

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181 Easy
181 of 995
Understanding +4 | -1
∫f(x)[f(x)]2dx=\int \frac{f(x)}{[f(x)]^2} d x=
View answer and explanation
Correct option: A
−[f(x)]−1+c-[f(x)]^{-1}+c
Option A (−[f(x)]−1+c-[f(x)]^{-1}+c) is correct because the integral of f′(x)[f(x)]2\frac{f'(x)}{[f(x)]^2} is found by using substitution, where f(x)=tf(x) = t.
182 Easy
182 of 995
Analyzing +4 | -1
∫ x2(ax  +  b)−2dx\int \,{x^2}{\left( {ax\,\, + \,\,b} \right)^{ - 2}}dx is equal to
View answer and explanation
Correct option: B
2a2  (x  −  ba  log⁡  (ax  +  b))  −  x2a  (ax  +  b)  +  c\frac{2}{{{a^2}}}\,\,\left( {x\,\, - \,\,\frac{b}{a}\,\,\log \,\,\left( {ax\,\, + \,\,b} \right)} \right)\,\, - \,\,\frac{{{x^2}}}{{a\,\,\left( {ax\,\, + \,\,b} \right)}}\,\, + \,\,c
Option B is correct because the integral is solved using substitution and integration by parts, resulting in the given expression.
183 Hard
183 of 995
Application +4 | -1
∫  x3(1+x2)1/3  dx\int \,\,\frac{{{x^3}}}{{{{\left( {1 + {x^2}} \right)}^{1/3}}}}\,\,dx is equal to
View answer and explanation
Correct option: B
320  (1+x2)2/3  (2x2−3)+C\frac{3}{{20}}\,\,{\left( {1 + {x^2}} \right)^{2/3}}\,\,\left( {2{x^2} - 3} \right) + C
Option B (320(1+x2)2/3(2x2−3)+C\frac{3}{{20}}\left( {1 + {x^2}} \right)^{2/3}\left( {2{x^2} - 3} \right) + C) is correct because it is the result of integrating the given function using substitution and integration by parts.
184 Easy
184 of 995
Application +4 | -1
∫tan⁡xsin⁡xcos⁡xdx=\int \frac{\sqrt{\tan x}}{\sin x \cos x} d x=
View answer and explanation
Correct option: B
2tan⁡x+c2 \sqrt{\tan x}+c
Option B is correct because the integral can be solved using substitution.
185 Easy
185 of 995
Application +4 | -1
The integral ∫dx(x+4)8/7(x−3)6/7\int {\frac{{dx}}{{{{\left( {x + 4} \right)}^{8/7}}{{\left( {x - 3} \right)}^{6/7}}}}} is equal to (where CC is a constant of integration)
View answer and explanation
Correct option: C
(x−3x+4)1/7+C{\left( {\frac{{x - 3}}{{x + 4}}} \right)^{1/7}} + C
Option C is correct because the integral is solved by substitution, resulting in the given expression plus a constant of integration.
186 Easy
186 of 995
Application +4 | -1
If ∫11  +  sin⁡  x  dx=tan⁡  (x2  +  a)  +  b\int \frac{1}{{1\,\, + \,\,\sin \,\,x}}\,\,dx = \tan \,\,\left( {\frac{x}{2}\,\, + \,\,a} \right)\,\, + \,\,b, then
View answer and explanation
Correct option: A
a=−π4,   b∈Ra = - \frac{\pi }{4},\,\;b \in R
Option A (a=−π4,  b∈Ra = - \frac{\pi }{4},\;b \in R) is correct because it is obtained by integrating the given expression using trigonometric identities and comparing it with the target form.
187 Easy
187 of 995
Application +4 | -1
∫ (1  +  x4)(1  −  x4)3/2    dx\int \,\frac{{\left( {1\,\, + \,\,{x^4}} \right)}}{{{{\left( {1\,\, - \,\,{x^4}} \right)}^{3/2}}}}\,\,\;dx is equal to
View answer and explanation
Correct option: A
x1  −  x4  +  c\frac{x}{{\sqrt {1\,\, - \,\,{x^4}} }}\,\, + \,\,c
Option A (x1−x4+c\frac{x}{{\sqrt {1 - {x^4}} }} + c) is correct because it is the result of integrating the given expression using substitution.
188 Easy
188 of 995
Understanding +4 | -1
Assertion : The value of the integral ∫ex[tan⁡x+sec⁡2x]dx\int e^x\left[\tan x+\sec ^2 x\right] d x is extan⁡x+C\mathrm{e}^{\mathrm{x}} \tan \mathrm{x}+\mathrm{C}

Reason: The value of the integral ex{f(x)+f′(x)}dxe^x\left\{f(x)+f^{\prime}(x)\right\} d x is exf(x)+C\mathrm{e}^{\mathrm{x}} \mathrm{f}(\mathrm{x})+\mathrm{C}.
View answer and explanation
Correct option: A
Assertion is correct, Reason is correct; Reason is a correct explanation for assertion.
Option A is correct because the assertion is true, and the reason correctly explains why the assertion is true. The integral of ex[f(x)+f′(x)]dxe^x[f(x) + f'(x)] dx is indeed exf(x)+Ce^x f(x) + C, and this applies to the given assertion with f(x)=tan⁡xf(x) = \tan x.
189 Medium
189 of 995
Application +4 | -1
The value of ∫0π∣sin⁡3  θ∣  dθ\int \nolimits_0^\pi |{\sin ^3}\,\,\theta |\,\,d\theta is
View answer and explanation
Correct option: C
43\frac{4}{3}
Option C is correct because the integral is evaluated by splitting the integral and using trigonometric substitution.
190 Easy
190 of 995
Application +4 | -1
∫(x2+x)(x−8+2x−9)1/10dx\int\left(x^2+x\right)\left(x^{-8}+2 x^{-9}\right)^{1 / 10} d x is equal to
View answer and explanation
Correct option: A
511(x2+2x)11/10+C\frac{5}{11}\left(x^2+2 x\right)^{11 / 10}+C
Option A is correct because the integral simplifies to 511(x2+2x)11/10+C\frac{5}{11}\left(x^2+2 x\right)^{11 / 10}+C after a suitable substitution.
191 Easy
191 of 995
Application +4 | -1
∫  sin⁡  x    dx3+4  cos⁡2  x\int \,\,\frac{{\sin \,\,x\;\,\,dx}}{{3 + 4\,\,{{\cos }^2}\,\,x}} is equal to
View answer and explanation
Correct option: C
−  12 3  tan⁡−1  (2  cos⁡  x3)  +  c- \,\,\frac{1}{{2\,\sqrt 3 }}\,\,{\tan ^{ - 1}}\,\,\left( {\frac{{2\,\,\cos \,\,x}}{{\sqrt 3 }}} \right)\,\, + \,\,c
Option C (− 123tan⁡−1(2cos⁡,x3)+,c- \,\frac{{1}}{{2\sqrt 3 }}\tan ^{ - 1}\left( {\frac{{2\cos \\,x}}{{\sqrt 3 }}} \right)\\ + \\,c) is the correct result of the integration after substitution and applying the standard integral formula.
192 Medium
192 of 995
Application +4 | -1
∫tan⁡−12x1−x2dx=\int \tan ^{-1} \frac{2 x}{1-x^2} d x=
View answer and explanation
Correct option: D
2xtan⁡−1x−log⁡(1+x2)+c2 x \tan ^{-1} x-\log \left(1+x^2\right)+c
Option D (2xtan⁡−1x−log⁡(1+x2)+c2 x \tan ^{-1} x-\log \left(1+x^2\right)+c) is correct because substituting x=tan⁡θx = \tan \theta simplifies the integral, and integration by parts yields the correct result.
193 Medium
193 of 995
Analyzing +4 | -1
∫x2+n(n−1)(xsin⁡x+ncos⁡x)2dx\int \frac{x^2+n(n-1)}{(x \sin x+n \cos x)^2} d x is equal to
View answer and explanation
Correct option: D
−xnsec⁡xxnsin⁡x+nxn−1cos⁡x+tan⁡x+c\frac{-x^n \sec x}{x^n \sin x+n x^{n-1} \cos x}+\tan x+c
Option B (−sec⁡xxsin⁡x+ncos⁡x+tan⁡x+c\frac{{-\sec x}}{{x \sin x+n \cos x}}+\tan x+c) is incorrect. Option D (−xnsec⁡xxnsin⁡x+nxn−1cos⁡x+tan⁡x+c\frac{{-x^n \sec x}}{{x^n \sin x+n x^{n-1} \cos x}}+\tan x+c) is the correct answer because after integration by parts and simplification, the expression matches this form.
194 Easy
194 of 995
Application +4 | -1
If ∫xx2−4x+8 dx=Klog⁡(x2−4x+8)+\int \frac{x}{x^2-4 x+8} \mathrm{~d} x=K \log \left(x^2-4 x+8\right)+ tan⁡−1(x−22)+C\tan ^{-1}\left(\frac{x-2}{2}\right)+C then the value of KK is
View answer and explanation
Correct option: A
1/21 / 2
Option A (1/2) is correct because it is the coefficient of the logarithmic term in the antiderivative after completing the square and using substitution.
195 Easy
195 of 995
Application +4 | -1
The value of ∫  2  dx1  −  4x2\int \,\,\frac{{2\,\,dx}}{{\sqrt {1\,\, - \,\,4{x^2}} }} is
View answer and explanation
Correct option: D
sin⁡−1  (2x)  +  c{\sin ^{ - 1}}\,\,\left( {2x} \right)\,\, + \,\,c
Option D (sin⁡−1  (2x) + c{\sin ^{ - 1}}\,\,\left( {2x} \right)\, + \,c) is correct because it is the direct result of integrating the given expression.
196 Medium
196 of 995
Analyzing +4 | -1
The value of the integral ∫1(x−3)6/7(x+4)8/7dx\int \frac{1}{(x-3)^{6 / 7}(x+4)^{8 / 7}} d x is
View answer and explanation
Correct option: A
(x−3x+4)1/7+C\left(\frac{x-3}{x+4}\right)^{1 / 7}+C
Option A is correct because after rewriting the integral and using a simple substitution, the integral evaluates to (x−3x+4)1/7+C(\frac{x-3}{x+4})^{1 / 7}+C.
197 Easy
197 of 995
Application +4 | -1
∫  1x    (log⁡ex  e)  dx\int \,\,\frac{1}{{x\;}}\,\,\left( {{{\log }_{ex}}\,\,e} \right)\,\,dx is equal to
View answer and explanation
Correct option: D
log⁡e  (1  +  log⁡e  x)  +  c{\log _e}\,\,(1\,\, + \,\,{\log _e}\,\,x)\,\, + \,\,c
Option D (log⁡e (1 + log⁡e x) + c{\log _e}\,(1\, + \,{\log _e}\,x)\, + \,c) is correct because after substituting log⁡ex=t{\log _e}x = t, the integral becomes a simple logarithmic form.
198 Easy
198 of 995
Application +4 | -1
If f(x)=x2−a2f(x)=\sqrt{x^2-a^2} then ∫x2f(x)dx\int \frac{x^2}{f(x)} \mathrm{d} x is
View answer and explanation
Correct option: D
x2f(x)+a22log⁡∣x+f(x)∣+C\frac{x}{2} f(x)+\frac{a^2}{2} \log |x+f(x)|+C
Option D (x2f(x)+a22log⁡∣x+f(x)∣+C\frac{x}{2} f(x)+\frac{a^2}{2} \log |x+f(x)|+C) is correct because it is the result of integrating the given function using integration by parts and trigonometric substitution.
199 Easy
199 of 995
Application +4 | -1
The integral ∫  dxx  (1+  log⁡  x)2\int \,\,\frac{{dx}}{{x\,\,{{\left( {1 + \,\,\log \,\,x} \right)}^2}}} is equal to
View answer and explanation
Correct option: B
−11  +  log⁡  x  +  c\frac{{ - 1}}{{1\,\, + \,\,\log \,\,x}}\,\, + \,\,c
Option B (−11  +  log⁡  x  +  c\frac{{ - 1}}{{1\,\, + \,\,\log \,\,x}}\,\, + \,\,c) is correct because it is obtained by using a simple u-substitution with u=1+log⁡xu = 1 + \log x.
200 Easy
200 of 995
Analyzing +4 | -1
∫1tan⁡x+cot⁡x+sec⁡x+cosec⁡xdx\int \frac{1}{\tan x+\cot x+\sec x+\operatorname{cosec} x} d x is equal to
View answer and explanation
Correct option: D
None of these
The correct answer is D (None of these) because the integral evaluates to a different expression than the given options.
201 Medium
201 of 995
Application +4 | -1
If f′(x)=tan⁡−1(sec⁡x+tan⁡x),x∈(−π2,π2)f^{\prime}(x)=\tan ^{-1}(\sec x+\tan x), x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) and f(0)=0f(0)=0, then the value of f(1)f(1) is
View answer and explanation
Correct option: A
π+14\frac{\pi+1}{4}
Option A is correct because the derivative simplifies to f′(x)=π4+x2f^{\prime}(x)=\frac{\pi}{4}+\frac{x}{2}, and integrating with the given condition f(0)=0f(0)=0 yields f(1)=π+14f(1)=\frac{\pi+1}{4}.
202 Easy
202 of 995
Application +4 | -1
∫e3log⁡x  (x4+1)−1  dx\int {e^{3\log x}}\,\,{\left( {{x^4} + 1} \right)^{ - 1}}\;dx equals
View answer and explanation
Correct option: B
14  log⁡  (x4+  1)+c\frac{1}{4}\,\,\log \,\,\left( {{x^4} + \,\,1} \right) + c
Option B (14  log⁡  (x4+  1)+c\frac{1}{4}\,\,\log \,\,\left( {{{{x^4}} + \,\,1}} \right) + c) is correct because it is the result of integrating the given expression after simplification and substitution.
203 Easy
203 of 995
Analyzing +4 | -1
Let f(x)=∫x2(1+x2)(1+1+x2)dxf(x)=\int \frac{x^2}{\left(1+x^2\right)\left(1+\sqrt{1+x^2}\right)} d x and f(0)=0f(0)=0. Then f(1)f(1) is equal to
View answer and explanation
Correct option: B
log⁡e(1+2)−π4\log _e(1+\sqrt{2})-\frac{\pi}{4}
Option B is correct because after substituting x=tan⁡θx = \tan \theta and simplifying, the integral evaluates to log⁡e(1+2)−π4\log _e(1+\sqrt{2})-\frac{\pi}{4}.
204 Easy
204 of 995
Application +4 | -1
∫x2x2+  4   dx\int \frac{{{x^2}}}{{{x^2} + \,\,4}}\,\;dx is equal to
View answer and explanation
Correct option: A
x  −  2 tan⁡−1  (x2)+cx\,\, - \,\,2\,{\tan ^{ - 1}}\,\,\left( {\frac{x}{2}} \right) + c
Option A (x−2tan⁡−1(x2)+cx - 2{\tan ^{ - 1}}(\frac{x}{2}) + c) is correct because after dividing and using the arctangent integral, this is the result.
205 Easy
205 of 995
Application +4 | -1
The value of ∫1sin⁡3xcos⁡5xdx\int \frac{1}{\sqrt{\sin ^3 x \cos ^5 x}} d x, is
View answer and explanation
Correct option: A
−2tan⁡x+23(tan⁡x)3/2+C\frac{-2}{\sqrt{\tan x}}+\frac{2}{3}(\tan x)^{3 / 2}+C
Option A (−2tan⁡x+23(tan⁡x)3/2+C\frac{{-2}}{{\sqrt{{\tan x}}}}+\frac{{2}}{{3}}(\tan x)^{{3 / 2}}+C) is correct because the integral is solved by manipulating the integrand and using substitution.
206 Easy
206 of 995
Application +4 | -1
The value of 2∫sin⁡x dxsin⁡(x−π4)\sqrt{2} \int \frac{\sin x \,d x}{\sin \left(x-\frac{\pi}{4}\right)} is
View answer and explanation
Correct option: B
x+log⁡∣sin⁡(x−π4)∣+Cx+\log \left|\sin \left(x-\frac{\pi}{4}\right)\right|+C
Option B (x+log⁡∣sin⁡(x−π4)∣+Cx+\log \left|\sin \left(x-\frac{{\pi}}{{4}}\right)\right|+C) is correct because it represents the result of the given integral after applying substitution and simplification.
207 Medium
207 of 995
Application +4 | -1
The value of ∫log⁡x(x+1)2dx\int \frac{\log x}{(x+1)^2} d x is
View answer and explanation
Correct option: A
−log⁡xx+1+log⁡x−log⁡(x+1)\frac{-\log x}{x+1}+\log x-\log (x+1)−log⁡xx+1+log⁡x−log⁡(x+1)\frac{-\log x}{x+1}+\log x-\log (x+1)
Option A is correct because the integral of log(x)/(x+1)2 dx is found using integration by parts, resulting in -log(x)/(x+1) + log(x) - log(x+1) + c.
208 Easy
208 of 995
Application +4 | -1
The value of ∫dxx+a2−x2\int \frac{d x}{x+\sqrt{a^2-x^2}}, is equal to
View answer and explanation
Correct option: A
12sin⁡−1(xa)+12log⁡∣x+a2−x2∣+C1\frac{1}{2} \sin ^{-1}\left(\frac{x}{a}\right)+\frac{1}{2} \log \left|x+\sqrt{a^2-x^2}\right|+C_1
Option A (12sin⁡−1(xa)+12log⁡∣x+a2−x2∣+C1\frac{{1}}{{2}} \sin ^{-1}(\frac{{x}}{{a}})+\frac{{1}}{{2}} \log |x+\sqrt{{a^2-x^2}}|+C_1) is correct because the integral evaluates to this expression after a trigonometric substitution and simplification.
209 Easy
209 of 995
Application +4 | -1
If In=∫sin⁡n  x dx{I_n} = \int {\sin ^n}\,\,x\,dx,then nIn  −  (n−1)In−2n{I_n}\,\, - \,\,\left( {n - 1} \right){I_{n - 2}} equals
View answer and explanation
Correct option: C
−  sin⁡n−1  x  cos⁡x- \,\,{\sin ^{n - 1}}\,\,x\,\,\cos x
Option C is correct because it is the result of applying integration by parts to the integral of sin^n(x).
210 Easy
210 of 995
Application +4 | -1
The value of ∫dx4x−3−x2\int \frac{d x}{\sqrt{4 x-3-x^2}} is equal to
View answer and explanation
Correct option: D
sin⁡−1(x−2)+C\sin ^{-1}(x-2)+C
Option D (sin⁡−1(x−2)+C\sin ^{-1}(x-2)+C) is correct because it is the result of completing the square and using the standard integral formula for arcsine.
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