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JEE • MATHEMATICS • INDEFINITE INTEGRATION

Indefinite integration practice · Easy

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91 Easy
91 of 746
Application +4 | -1
∫1xsin⁡xdx=\int \frac{1}{\sqrt{x}} \sin \sqrt{x} d x=
View answer and explanation
Correct option: B
−2cos⁡x+c-2 \cos \sqrt{x}+c
Option B (−2cos⁡x+c-2 \cos \sqrt{x}+c) is correct because the integral evaluates to this after a simple substitution.
92 Easy
92 of 746
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.
93 Easy
93 of 746
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.
94 Easy
94 of 746
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.
95 Easy
95 of 746
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.
96 Easy
96 of 746
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}.
97 Easy
97 of 746
Application +4 | -1
If ∫cot⁡xsin⁡xcos⁡xdx=Acot⁡x+B\int \frac{\sqrt{\cot x}}{\sin x \cos x} d x=A \sqrt{\cot x}+B, then AA is equal to
View answer and explanation
Correct option: D
−2-2
Option D (-2) is correct because after simplifying the integral and performing a u-substitution, the coefficient of cot⁡x\sqrt{\cot x} is -2.
98 Easy
98 of 746
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).
99 Easy
99 of 746
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.
100 Easy
100 of 746
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.
101 Easy
101 of 746
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.
102 Easy
102 of 746
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.
103 Easy
103 of 746
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.
104 Easy
104 of 746
Application +4 | -1
Value of ∫2810−xx+10−xdx\int\limits_{2}^8 \frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}} d x is
View answer and explanation
Correct option: A
3
Option A (3) is correct because the definite integral can be solved by using the property ∫abf(x)dx=∫abf(a+b−x)dx\int_a^b f(x) dx = \int_a^b f(a+b-x) dx and adding the two integrals.
105 Easy
105 of 746
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.

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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
106 Easy
106 of 746
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.
107 Easy
107 of 746
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.
108 Easy
108 of 746
Application +4 | -1
∫dx(1+x2)p2+q2(tan⁡−1x)2=\int \frac{d x}{\left(1+x^2\right) \sqrt{p^2+q^2\left(\tan ^{-1} x\right)^2}}=
View answer and explanation
Correct option: A
1qlog⁡[qtan⁡−1x+p2+q2(tan⁡−1x)2]+c\frac{1}{q} \log \left[q \tan ^{-1} x+\sqrt{p^2+q^2\left(\tan ^{-1} x\right)^2}\right]+c
Option A is correct because the integral can be solved by substituting qtan⁡−1x=tq \tan^{-1} x = t and using the standard integral formula.
109 Easy
109 of 746
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.
110 Easy
110 of 746
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.
111 Easy
111 of 746
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.
112 Easy
112 of 746
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.
113 Easy
113 of 746
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.
114 Easy
114 of 746
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.
115 Easy
115 of 746
Analyzing +4 | -1
∫e(x2+4ln⁡x)−x3ex2x−1dx\int \frac{e^{\left(x^2+4 \ln x\right)}-x^3 e^{x^2}}{x-1} d x equals to
View answer and explanation
Correct option: D
None of these
The correct answer is D (None of these) because the result of the integration is 12(x2−1)ex2+C\frac{{1}}{{2}}(x^2-1) e^{x^2}+C, which does not match any of the provided options.
116 Easy
116 of 746
Application +4 | -1
∫d(ln⁡x)x+1=\int \frac{d(\ln x)}{x+1}=
View answer and explanation
Correct option: C
ln⁡∣xx+1∣+c\ln \left|\frac{x}{x+1}\right|+c
Option C, ln⁡∣xx+1∣+c\ln \left|\frac{{x}}{{x+1}}\right|+c, is correct because the integral of d(ln⁡x)x+1\frac{{d(\ln x)}}{{x+1}} can be solved by substitution, leading to this result.
117 Easy
117 of 746
Application +4 | -1
∫32  x3(log⁡  x)2dx\int 32\,\,{x^3}{\left( {\log \,\,x} \right)^2}dx is equal to
View answer and explanation
Correct option: B
x4  { 8(log⁡  x)2− 4  log⁡  x+1 }+c{x^4}\,\,\left\{ {\,8{{\left( {\log \,\,x} \right)}^2} - \,4\,\,\log \,\,x + 1\,} \right\} + c
Option B (x4  { 8(log⁡ x)2− 4  log⁡ x+1 }+c{x^4}\,\,\{\,8{{\left( {\log \,x} \right)}^2} - \,4\,\,\log \,x + 1\,\} + c) is correct because it is the result of applying integration by parts twice to the given integral.
118 Easy
118 of 746
Application +4 | -1
∫ex(1+tan⁡x)sec⁡xdx=\int e^x(1+\tan x) \sec x d x=
View answer and explanation
Correct option: C
exsec⁡xe^x \sec x
Option C is correct because the integral can be simplified to the form ∫ex[f(x)+f′(x)]dx\int e^x[f(x) + f'(x)] dx, where f(x)=sec⁡xf(x) = \sec x and f′(x)=sec⁡xtan⁡xf'(x) = \sec x \tan x.
119 Easy
119 of 746
Application +4 | -1
∫x3ex2dx=\int x^3 e^{x^2} d x=
View answer and explanation
Correct option: C
12(x2−1)ex2+c\frac{1}{2}\left(x^2-1\right) e^{x^2}+c
Option C is correct because the integral of x3 e(x2) dx is found using substitution and integration by parts, resulting in 1/2(x2-1)e(x2) + c.
120 Easy
120 of 746
Application +4 | -1
∫dx(x2+1)(x2+4)=\int \frac{d x}{\left(x^2+1\right)\left(x^2+4\right)}=
View answer and explanation
Correct option: C
13tan⁡−1x−16tan⁡−1x2+c\frac{1}{3} \tan ^{-1} x-\frac{1}{6} \tan ^{-1} \frac{x}{2}+c
Option C, 13tan⁡−1x−16tan⁡−1x2+c\frac{1}{3} \tan ^{-1} x-\frac{1}{6} \tan ^{-1} \frac{x}{2}+c, is the correct answer because it results from partial fraction decomposition and integration of the given expression.
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