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JEE • MATHEMATICS • INDEFINITE INTEGRATION

Indefinite integration practice · Mixed difficulty

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121 Medium
121 of 995
Application +4 | -1
∫1−x1+xdx=\int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} d x=
View answer and explanation
Correct option: A
−21−x+Cos⁡−1x+x−x2+c-2 \sqrt{1-x}+\operatorname{Cos}^{-1} \sqrt{x}+\sqrt{x-x^2}+c
Option A (−21−x+Cos⁡−1x+x−x2+c-2 \sqrt{1-x}+\operatorname{Cos}^{-1} \sqrt{x}+\sqrt{x-x^2}+c) is correct because it is the result of trigonometric substitution and simplification of the integral.
122 Easy
122 of 995
Application +4 | -1
∫xexdx\int x e^x d x is equal to
View answer and explanation
Correct option: B
(x−1)ex+c(x-1) e^x+c
Option B is correct because the integral of x e^x dx is found using integration by parts, resulting in (x-1)e^x + c.
123 Easy
123 of 995
Application +4 | -1
The value of ∫  x2+1x4−x2+1  dx\int \,\,\frac{{{x^2} + 1}}{{{x^4} - {x^2} + 1}}\,\,dx is
View answer and explanation
Correct option: D
tan⁡−1  (x2  −  1x)  +  c{\tan ^{ - 1}}\,\,\left( {\frac{{{x^2}\,\, - \,\,1}}{x}} \right)\,\, + \,\,c
Option D is correct because the integral of (x2 + 1)/(x4 - x2 + 1) is solved by dividing by x2 and using a substitution, resulting in arctan((x2 - 1)/x) + c.
124 Medium
124 of 995
Application +4 | -1
∫log⁡xx2dx\int \frac{\log x}{x^2} d x is equal to
View answer and explanation
Correct option: D
−(1+log⁡x)x+c-\frac{(1+\log x)}{x}+c
Option D (-(1+log⁡x)x+c\frac{(1+\log x)}{x}+c) is correct because by using integration by parts, we can evaluate the integral.
125 Easy
125 of 995
Analyzing +4 | -1
Let I(x)=∫(x+1)x(1+xex)2dx,x>0I(x)=\int \frac{(x+1)}{x\left(1+x e^{x}\right)^{2}} d x, x > 0. If lim⁡x→∞I(x)=0\lim \limits_{{x \rightarrow \infty}} I(x)=0, then I(1)I(1) is equal to
View answer and explanation
Correct option: A
e+2e+1−log⁡e(e+1)\frac{e+2}{e+1}-\log _{e}(e+1)
Option B (x2+7x+12=0x^{2}+7 x+12=0) is correct because solving the given equation for z yields values for alpha and beta, whose sum and product are the roots of the equation x2+7x+12=0x^{2}+7 x+12=0.
126 Medium
126 of 995
Analyzing +4 | -1
The anti-derivative of f  (x)=log⁡  (log⁡  x)  +  (log⁡  x)−2f\,\,\left( x \right) = \log \,\,(\log \,\,x)\,\, + \,\,{(\log \,\,x)^{ - 2}} whose graph passes through (e,  e)\left( {e,\,\,e} \right), is
View answer and explanation
Correct option: C
x [log⁡  (log⁡  x)  −  (log⁡  x)−1]  +  2ex\,[\log \,\,(\log \,\,x)\,\, - \,\,{(\log \,\,x)^{ - 1}}]\,\, + \,\,2e
Option C (x [log⁡ (log⁡ x) −  (log⁡ x)−1] + 2ex\,[\log \,(\log \,x)\, - \,\,{(\log \,x)^{ - 1}}]\, + \,2e) is correct because it is the antiderivative of the given function that passes through the point (e, e).
127 Easy
127 of 995
Application +4 | -1
∫ex(x+1)cos⁡2(xex)dx=\int \frac{e^x(x+1)}{\cos ^2\left(x e^x\right)} d x=
View answer and explanation
Correct option: A
tan⁡(xex)+c\tan \left(x e^x\right)+c
Option A is correct because the integral can be solved by using a simple substitution.
128 Medium
128 of 995
Application +4 | -1
∫ex(1−sin⁡x1−cos⁡x)dx\int e^x\left(\frac{1-\sin x}{1-\cos x}\right) d x is equal to
View answer and explanation
Correct option: B
−excot⁡(x/2)-e^x \cot (x / 2)
Option B is correct because the integral simplifies to −excot⁡(x/2)+c-e^x \cot(x/2) + c by rewriting the integrand and 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.
129 Easy
129 of 995
Application +4 | -1
∫ex[1+xlog⁡xx]dx=\int e^x\left[\frac{1+x \log x}{x}\right] d x=
View answer and explanation
Correct option: D
exlog⁡x+ce^x \log x+c
Option D is correct because the integral can be expressed in the form ∫ex[f(x)+f′(x)]dx\int e^x [f(x) + f'(x)] dx, where f(x)=log⁡xf(x) = \log x and f′(x)=1xf'(x) = \frac{1}{x}.
130 Medium
130 of 995
Application +4 | -1
∫sin⁡(log⁡x)dx=\int \sin (\log x) d x=
View answer and explanation
Correct option: C
12x[sin⁡(log⁡x)−cos⁡(log⁡x)]\frac{1}{2} x[\sin (\log x)-\cos (\log x)]
Option C (12x[sin⁡(log⁡x)−cos⁡(log⁡x)]\frac{1}{2} x[\sin (\log x)-\cos (\log x)]) is correct because it is the result of integrating sin⁡(log⁡x)\sin (\log x) using integration by parts twice.
131 Easy
131 of 995
Application +4 | -1
∫dxex+e−x=\int \frac{d x}{e^x+e^{-x}}=
View answer and explanation
Correct option: B
tan⁡−1(ex)\tan ^{-1}\left(e^x\right)
Option B is correct because the integral can be rewritten and solved using a substitution, resulting in tan⁡−1(ex)\tan^{-1}(e^x).
132 Easy
132 of 995
Application +4 | -1
∫0∞6e3x+6e2x+11ex+6dx=\int\limits_{{0}}^{\infty} \frac{6}{e^{3 x}+6 e^{2 x}+11 e^{x}+6} d x=
View answer and explanation
Correct option: A
log⁡e(3227)\log _{e}\left(\frac{32}{27}\right)
Option A (log⁡e(3227)\log_e(\frac{32}{27})) is correct because it is the result of evaluating the definite integral after performing partial fraction decomposition.
133 Easy
133 of 995
Analyzing +4 | -1
∫x2(1−ln⁡x)(ln⁡x)4−x4dx\int \frac{x^2(1-\ln x)}{(\ln x)^4-x^4} d x is equal to
View answer and explanation
Correct option: B
14ln⁡(ln⁡x−xln⁡x+x)−12tan⁡−1(ln⁡xx)+C\frac{1}{4} \ln \left(\frac{\ln x-x}{\ln x+x}\right)-\frac{1}{2} \tan ^{-1}\left(\frac{\ln x}{x}\right)+C
Option D is correct because after substitution and partial fraction decomposition, the integral evaluates to the expression in option D.
134 Medium
134 of 995
Application +4 | -1
∫  cos⁡  [2  cot⁡−1  1−x1+x]  dx\int \,\,\cos \,\,\left[ {2\,\,{{\cot }^{ - 1}}\,\,\sqrt {\frac{{1 - x}}{{1 + x}}} } \right]\,\,dx is equal to
View answer and explanation
Correct option: C
−  12  x2  +  c- \,\,\frac{1}{2}\,\,{x^2}\,\, + \,\,c
Option C ( −  12 x2 + c\ - \,\,\frac{1}{2}\,{x^2}\, + \,c) is correct. Simplifying the expression inside the integral using trigonometric identities leads to the result.
135 Easy
135 of 995
Understanding +4 | -1
∫  xe−1  +  ex−1xe  +  ex    dx\int \,\,\frac{{{x^{e - 1}}\,\, + \,\,{e^{x - 1}}}}{{{x^e}\,\, + \,\,{e^x}}}\;\,\,dx is equal to
View answer and explanation
Correct option: C
1e  log⁡  (xe  +  ex)+c\frac{1}{e}\,\,\log \,\,({x^e}\,\, + \,\,{e^x}) + c
Option C, 1e log⁡ (xe + ex)+c\frac{1}{e}\,\log \,({x^e}\,+\,{e^x}) + c, is correct because it results from a u-substitution where the derivative of the denominator is proportional to the numerator.

Explore Other Chapters in MATHEMATICS

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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
136 Easy
136 of 995
Application +4 | -1
f(x)=∫dxsin⁡4xf(x)=\int \frac{d x}{\sin ^4 x} is a
View answer and explanation
Correct option: A
polynomial of degree 3 in cot⁡x\cot x
The correct answer is A because after rewriting the integrand in terms of cot⁡x\cot x and integrating, the result is a polynomial of degree 3 in cot⁡x\cot x.
137 Medium
137 of 995
Application +4 | -1
∫sin⁡−1x(1−x2)3/2dx=\int \frac{\sin ^{-1} x}{\left(1-x^2\right)^{3 / 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 \left(1-x^2\right)+c) is correct because substituting sin⁡t=x\sin t = x simplifies the integral, and integration by parts yields the correct result.
138 Easy
138 of 995
Understanding +4 | -1
∫sec⁡pxtan⁡xdx=\int \sec ^p x \tan x d x=
View answer and explanation
Correct option: B
sec⁡pxp+c\frac{\sec ^p x}{p}+c
Option B (sec⁡pxp+c\frac{\sec ^p x}{p}+c) is correct because the derivative of sec⁡x\sec x is sec⁡xtan⁡x\sec x \tan x, which allows for a direct substitution.
139 Easy
139 of 995
Application +4 | -1
∫f′(x)f  (x)  log⁡  [f(x)]  dx\int \frac{{f'\left( x \right)}}{{f\,\,\left( x \right)\,\,\log \,\,\left[ {f\left( x \right)} \right]}}\,\,dx is equal to
View answer and explanation
Correct option: C
log⁡  [ log⁡  f(x) ]  +  c\log \,\,[\,\log \,\,f\left( x \right)\,]\,\, + \,\,c
Option C is correct because the integral is solved by recognizing the form and applying a logarithmic substitution.
140 Easy
140 of 995
Application +4 | -1
∫sec⁡xtan⁡3xdx=\int \sec x \tan ^3 x d x=
View answer and explanation
Correct option: A
13sec⁡3x−sec⁡x+c\frac{1}{3} \sec ^3 x-\sec x+c
Option A is correct because the integral is solved by using trigonometric identities and substitution, resulting in 13sec⁡3x−sec⁡x+c\frac{1}{3} \sec ^3 x-\sec x+c.
141 Easy
141 of 995
Analyzing +4 | -1
If f(x)=∫5x8+7x6(x2+1+2x7)2dx,x≥0f(x)=\int \frac{5 x^8+7 x^6}{\left(x^2+1+2 x^7\right)^2} d x, x \geq 0 and f(0)=0f(0)=0, then the value of f(1)f(1) is
View answer and explanation
Correct option: D
14\frac{1}{4}
Option D (1/4) is correct because after finding the indefinite integral and applying the initial condition f(0)=0, we evaluate f(1) to get 1/4.
142 Easy
142 of 995
Application +4 | -1
The value of ∫xlog⁡x(log⁡x−1)dx\int x \log x(\log x-1) d x is equal to
View answer and explanation
Correct option: B
12(xlog⁡x−x)2+C\frac{1}{2}(x \log x-x)^2+C
Option B is correct because the integral can be simplified to 12(xlog⁡x−x)2+C\frac{1}{2}(x \log x - x)^2 + C by recognizing the derivative pattern.
143 Medium
143 of 995
Understanding +4 | -1
If ∫2x+3x2−5x+6dx=9ln⁡(x−3)−7ln⁡(x−2)+A\int \frac{2 x+3}{x^2-5 x+6} d x=9 \ln (x-3)-7 \ln (x-2)+A then A=A=
View answer and explanation
Correct option: C
Constant
Option C (Constant) is correct because after integrating the given function and comparing it with the provided form, the remaining term is a constant.
144 Easy
144 of 995
Application +4 | -1
∫  ex(2+ex)  (ex+1)  dx\int \,\,\frac{{{e^x}}}{{\left( {2 + {e^x}} \right)\,\,\left( {{e^x} + 1} \right)}}\,\,dx
is equal to
View answer and explanation
Correct option: A
log⁡  (ex+  1ex+  2)  +  c\log \,\,\left( {\frac{{{e^x} + \,\,1}}{{{e^x} + \,\,2}}} \right)\,\, + \,\,c
Option A (log((e^x + 1)/(e^x + 2)) + c) is correct because by using substitution and partial fraction decomposition, the integral can be solved to obtain the given result.
145 Medium
145 of 995
Analyzing +4 | -1
∫x4(1−x)41+x2dx=f(x)−4Tan⁡−1x+c\int \frac{x^4(1-x)^4}{1+x^2} d x=f(x)-4 \operatorname{Tan}^{-1} x+c where f(x)f(x) is
View answer and explanation
Correct option: A
x77−23x6+x5−4x33+4x\frac{x^7}{7}-\frac{2}{3} x^6+x^5-\frac{4 x^3}{3}+4 x
The correct answer is A (x77−23x6+x5−4x33+4x\frac{x^7}{7}-\frac{2}{3} x^6+x^5-\frac{4 x^3}{3}+4 x) because after polynomial long division, the integral is straightforward and the result matches option A.
146 Easy
146 of 995
Understanding +4 | -1
∫{ f(x)  g′′ (x)  −  f′′  (x)  g(x)}  dx\int \left\{ {\,f\left( x \right)\,\,g''\,\left( x \right)\,\, - \,\,f''\,\,\left( x \right)\,\,g\left( x \right)} \right\}\,\,dx is equal to
View answer and explanation
Correct option: C
f(x) g′  (x)  −  f′(x)g′  (x)f\left( x \right)\,g'\,\,\left( x \right)\,\, - \,\,f'\left( x \right)g'\,\,\left( x \right)
Option B (f'(x)g(x) - f(x)g'(x)) is correct because it results from applying integration by parts twice to the given expression.
147 Easy
147 of 995
Application +4 | -1
If ∫12ax−x2dx=fog(x)+C\int \frac{1}{\sqrt{2 a x-x^2}} d x=fog(x)+C, then
View answer and explanation
Correct option: B
f(x)=sin⁡−1xf(x)=\sin ^{-1} x, and g(x)=x−aag(x)=\frac{x-a}{a}
Option B, f(x)=sin⁡−1xf(x)=\sin ^{-1} x, and g(x)=x−aag(x)=\frac{x-a}{a}, is correct because completing the square and using a trigonometric substitution leads to this result.
148 Easy
148 of 995
Application +4 | -1
∫sec⁡2x1+tan⁡xdx=\int \frac{\sec ^2 x}{1+\tan x} d x=
View answer and explanation
Correct option: C
log⁡(1+tan⁡x)+c\log (1+\tan x)+c
Option C is correct because the integral can be solved using substitution.
149 Medium
149 of 995
Analyzing +4 | -1
If In=∫(sin⁡x+cos⁡x)ndx(n≥2)I_n=\int(\sin x+\cos x)^n d x(n \geq 2), then nIn−2(n−1)In−2n I_n-2(n-1) I_{n-2} is equal to
View answer and explanation
Correct option: B
(sin⁡x+cos⁡x)n−1(sin⁡x−cos⁡x)(\sin x+\cos x)^{n-1}(\sin x-\cos x)
Option B ((sin⁡x+cos⁡x)n−1(sin⁡x−cos⁡x)\sin x+\cos x)^{n-1}(\sin x-\cos x)) is correct because after integration by parts and simplification, the expression matches this form.
150 Easy
150 of 995
Application +4 | -1
∫1cos⁡x−sin⁡xdx\int \frac{1}{\cos x-\sin x} d x is equal to
View answer and explanation
Correct option: D
12log⁡∣tan⁡(x2+3π8)∣+C\frac{1}{\sqrt{2}} \log \left|\tan \left(\frac{x}{2}+\frac{3 \pi}{8}\right)\right|+C
Option D, 12log⁡∣tan⁡(x2+3π8)∣+C\frac{1}{\sqrt{2}} \log \left|\tan \left(\frac{x}{2}+\frac{3 \pi}{8}\right)\right|+C, is correct because rewriting the integrand using trigonometric identities and integrating leads to this result.
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