Integration by Substitution

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jee-main 2014 Q83 Substitution to Transform Integral Form (Show Transformed Expression)
The integral $\int \left( 1 + x - \frac { 1 } { x } \right) e ^ { x + \frac { 1 } { x } } d x$, is equal to
(1) $( x + 1 ) e ^ { x + \frac { 1 } { x } } + c$
(2) $- x e ^ { x + \frac { 1 } { x } } + c$
(3) $( x - 1 ) e ^ { x + \frac { 1 } { x } } + c$
(4) $x e ^ { x + \frac { 1 } { x } } + c$
jee-main 2014 Q83 Substitution to Transform Integral Form (Show Transformed Expression)
If $m$ is a non-zero number and $\int \frac { x ^ { 5 m - 1 } + 2 x ^ { 4 m - 1 } } { \left( x ^ { 2 m } + x ^ { m } + 1 \right) ^ { 3 } } d x = f ( x ) + c$, then $f ( x )$ is equal to
(1) $\frac { \left( x ^ { 5 m } - x ^ { 4 m } \right) } { 2 m \left( x ^ { 2 m } + x ^ { m } + 1 \right) ^ { 2 } }$
(2) $\frac { 1 } { 2 m } \frac { x ^ { 4 m } } { \left( x ^ { 2 m } + x ^ { m } + 1 \right) ^ { 2 } }$
(3) $\frac { x ^ { 5 m } } { 2 m \left( x ^ { 2 m } + x ^ { m } + 1 \right) ^ { 2 } }$
(4) $\frac { 2 m \left( x ^ { 5 m } + x ^ { 4 m } \right) } { \left( x ^ { 2 m } + x ^ { m } + 1 \right) ^ { 2 } }$
jee-main 2014 Q84 Substitution to Evaluate a Definite Integral (Numerical Answer)
The integral $\int _ { 0 } ^ { \frac { 1 } { 2 } } \frac { \ln ( 1 + 2 x ) } { 1 + 4 x ^ { 2 } } d x$ equals
(1) $\frac { \pi } { 4 } \ln 2$
(2) $\frac { \pi } { 16 } \ln 2$
(3) $\frac { \pi } { 8 } \ln 2$
(4) $\frac { \pi } { 32 } \ln 2$
jee-main 2014 Q84 Substitution to Compute an Indefinite Integral with Initial Condition
Let, the function $F$ be defined as $F ( x ) = \int _ { 1 } ^ { x } \frac { e ^ { t } } { t } d t , x > 0$, then the value of the integral $\int _ { 1 } ^ { x } \frac { e ^ { t } } { t + a } d t$, where $a > 0$, is
(1) $e ^ { a } [ F ( x ) - F ( 1 + a ) ]$
(2) $e ^ { - a } [ F ( x + a ) - F ( a ) ]$
(3) $e ^ { a } [ F ( x + a ) - F ( 1 + a ) ]$
(4) $e ^ { - a } [ F ( x + a ) - F ( 1 + a ) ]$
jee-main 2015 Q69 Substitution to Compute an Indefinite Integral with Initial Condition
The integral $\int \frac{dx}{x^2(x^4+1)^{3/4}}$ equals:
(1) $-\left(1 + \frac{1}{x^4}\right)^{1/4} + C$
(2) $-\left(\frac{x^4+1}{x^4}\right)^{1/4} + C$
(3) $\left(\frac{x^4+1}{x^4}\right)^{1/4} + C$
(4) $\left(1 + \frac{1}{x^4}\right)^{1/4} + C$
jee-main 2015 Q83 Substitution to Transform Integral Form (Show Transformed Expression)
The integral $\int \frac { d x } { x ^ { 2 } \left( x ^ { 4 } + 1 \right) ^ { \frac { 3 } { 4 } } }$ equals to
(1) $- \left( \frac { x ^ { 4 } + 1 } { x ^ { 4 } } \right) ^ { \frac { 1 } { 4 } } + c$
(2) $\left( \frac { x ^ { 4 } + 1 } { x ^ { 4 } } \right) ^ { \frac { 1 } { 4 } } + c$
(3) $\left( x ^ { 4 } + 1 \right) ^ { \frac { 1 } { 4 } } + c$
(4) $- \left( x ^ { 4 } + 1 \right) ^ { \frac { 1 } { 4 } } + c$
jee-main 2016 Q73 Substitution to Compute an Indefinite Integral with Initial Condition
The integral $\int \frac{2x^{12} + 5x^9}{(x^5 + x^3 + 1)^3} dx$ is equal to: (1) $\frac{-x^{10}}{2(x^5+x^3+1)^2} + C$ (2) $\frac{x^{10}}{2(x^5+x^3+1)^2} + C$ (3) $\frac{-x^5}{(x^5+x^3+1)^2} + C$ (4) $\frac{x^5}{2(x^5+x^3+1)^2} + C$
jee-main 2016 Q83 Substitution to Transform Integral Form (Show Transformed Expression)
The integral $\int \frac { d x } { ( 1 + \sqrt { x } ) \sqrt { x - x ^ { 2 } } }$ is equal to
(1) $- 2 \sqrt { \frac { 1 + \sqrt { x } } { 1 - \sqrt { x } } } + c$
(2) $- \sqrt { \frac { 1 - \sqrt { x } } { 1 + \sqrt { x } } } + c$
(3) $- 2 \sqrt { \frac { 1 - \sqrt { x } } { 1 + \sqrt { x } } } + c$
(4) $\sqrt { \frac { 1 + \sqrt { x } } { 1 - \sqrt { x } } } + c$
jee-main 2016 Q85 Substitution to Compute an Indefinite Integral with Initial Condition
The integral $\int \frac{2x^{12} + 5x^9}{(x^5 + x^3 + 1)^3} dx$ is equal to:
(1) $\frac{-x^{10}}{2(x^5 + x^3 + 1)^2} + C$
(2) $\frac{x^{10}}{2(x^5 + x^3 + 1)^2} + C$
(3) $\frac{-x^5}{(x^5 + x^3 + 1)^2} + C$
(4) $\frac{x^5}{2(x^5 + x^3 + 1)^2} + C$
jee-main 2016 Q86 Substitution to Evaluate a Definite Integral (Numerical Answer)
The integral $\int_0^{\pi/4} \frac{\sin x + \cos x}{9 + 16\sin 2x} dx$ is equal to:
(1) $\frac{1}{20} \log 3$
(2) $\log 3$
(3) $\frac{1}{20} \log 9$
(4) $\frac{1}{10} \log 3$
jee-main 2016 Q88 Substitution to Compute an Indefinite Integral with Initial Condition
If $m$ is a non-zero number and $\int \frac{x^{5m-1} + 2x^{4m-1}}{(x^{2m} + x^m + 1)^3} dx = f(x) + C$, then $f(x)$ is:
(1) $\frac{x^{5m}}{2m(x^{2m} + x^m + 1)^2}$
(2) $\frac{x^{4m}}{2m(x^{2m} + x^m + 1)^2}$
(3) $\frac{(2x^{2m} + x^m)}{(x^{2m} + x^m + 1)^2}$
(4) $\frac{(x^{5m} + x^{4m})}{2m(x^{2m} + x^m + 1)^2}$
jee-main 2018 Q83 Substitution to Transform Integral Form (Show Transformed Expression)
The integral $\int \frac { \sin ^ { 2 } x \cos ^ { 2 } x } { \left( \sin ^ { 5 } x + \cos ^ { 3 } x \sin ^ { 2 } x + \sin ^ { 3 } x \cos ^ { 2 } x + \cos ^ { 5 } x \right) ^ { 2 } } d x$, is equal to (where $C$ is the constant of integration).
(1) $\frac { - 1 } { 1 + \cot ^ { 3 } x } + C$
(2) $\frac { 1 } { 3 \left( 1 + \tan ^ { 3 } x \right) } + C$
(3) $\frac { - 1 } { 3 \left( 1 + \tan ^ { 3 } x \right) } + C$
(4) $\frac { 1 } { 1 + \cot ^ { 3 } x } + C$
jee-main 2018 Q83 Substitution to Transform Integral Form (Show Transformed Expression)
If $f \left( \frac { x - 4 } { x + 2 } \right) = 2 x + 1 , ( x \in R - \{ 1 , - 2 \} )$, then $\int f ( x ) d x$ is equal to
(1) $12 \ln | 1 - x | - 3 x + C$
(2) $- 12 \ln | 1 - x | - 3 x + C$
(3) $12 \ln | 1 - x | + 3 x + C$
(4) $- 12 \ln | 1 - x | + 3 x + C$
jee-main 2018 Q83 Substitution to Transform Integral Form (Show Transformed Expression)
If $f \left( \frac { x - 4 } { x + 2 } \right) = 2 x + 1 , ( x \in R = \{ 1 , - 2 \} )$, then $\int f ( x ) d x$ is equal to (where $C$ is a constant of integration)
(1) $12 \log _ { e } | 1 - x | - 3 x + c$
(2) $- 12 \log _ { e } | 1 - x | - 3 x + c$
(3) $- 12 \log _ { e } | 1 - x | + 3 x + c$
(4) $12 \log _ { e } | 1 - x | + 3 x + c$
jee-main 2019 Q82 Substitution to Transform Integral Form (Show Transformed Expression)
The integral $\int \frac { 3 x ^ { 13 } + 2 x ^ { 11 } } { \left( 2 x ^ { 4 } + 3 x ^ { 2 } + 1 \right) ^ { 4 } } d x$, is equal to
(1) $\frac { x ^ { 4 } } { 6 \left( 2 x ^ { 4 } + 3 x ^ { 2 } + 1 \right) ^ { 3 } } + C$
(2) $\frac { x ^ { 4 } } { \left( 2 x ^ { 4 } + 3 x ^ { 2 } + 1 \right) ^ { 3 } } + C$
(3) $\frac { x ^ { 12 } } { \left( 2 x ^ { 4 } + 3 x ^ { 2 } + 1 \right) ^ { 3 } } + C$
(4) $\frac { x ^ { 12 } } { 6 \left( 2 x ^ { 4 } + 3 x ^ { 2 } + 1 \right) ^ { 3 } } + C$
jee-main 2019 Q83 Substitution to Transform Integral Form (Show Transformed Expression)
Let, $n \geq 2$ be a natural number and $0 < \theta < \frac { \pi } { 2 }$. Then $\int \frac { \left( \sin ^ { n } \theta - \sin \theta \right) ^ { \frac { 1 } { n } } \cos \theta } { \sin ^ { n + 1 } \theta } d \theta$, is equal to
(1) $\frac { n } { n ^ { 2 } - 1 } \left( 1 - \frac { 1 } { \sin ^ { n + 1 } \theta } \right) ^ { \frac { n + 1 } { n } } + c$
(2) $\frac { n } { n ^ { 2 } + 1 } \left( 1 - \frac { 1 } { \sin ^ { n - 1 } \theta } \right) ^ { \frac { n + 1 } { n } } + c$
(3) $\frac { n } { n ^ { 2 } - 1 } \left( 1 - \frac { 1 } { \sin ^ { n - 1 } \theta } \right) ^ { \frac { n + 1 } { n } } + c$
(4) $\frac { n } { n ^ { 2 } - 1 } \left( 1 + \frac { 1 } { \sin ^ { n - 1 } \theta } \right) ^ { \frac { n + 1 } { n } } + c$
jee-main 2019 Q83 Substitution to Transform Integral Form (Show Transformed Expression)
$\int \sec ^ { 2 } x \cdot \cot ^ { \frac { 4 } { 3 } } x \, d x$ is equal to
(1) $3 \tan ^ { - \frac { 1 } { 3 } } x + C$
(2) $- \frac { 3 } { 4 } \tan ^ { - \frac { 4 } { 3 } } x + C$
(3) $- 3 \tan ^ { - \frac { 1 } { 3 } } x + C$
(4) $- 3 \cot ^ { - \frac { 1 } { 3 } } x + C$
jee-main 2019 Q84 Substitution to Compute an Indefinite Integral with Initial Condition
If $f(x) = \int \frac{\left(5x^8 + 7x^6\right)}{\left(x^2 + 1 + 2x^7\right)^2}\,dx,\,(x \geq 0)$, and $f(0) = 0$, then the value of $f(1)$ is
(1) $\frac{-1}{4}$
(2) $\frac{1}{2}$
(3) $\frac{1}{4}$
(4) $-\frac{1}{2}$
jee-main 2019 Q84 Substitution to Transform Integral Form (Show Transformed Expression)
If $\int \frac { d x } { x ^ { 3 } \left( 1 + x ^ { 6} \right) ^ { \frac { 2 } { 3 } } } = x f ( x ) \left( 1 + x ^ { 6} \right)^{ \frac { 1 } { 3 } } + C$, where $C$ is a constant of integration, then the function $f ( x )$ is equal to
(1) $\frac { 3 } { x ^ { 2 } }$
(2) $- \frac { 1 } { 2 x ^ { 3 } }$
(3) $- \frac { 1 } { 6 x ^ { 3 } }$
(4) $- \frac { 1 } { 2 x ^ { 2 } }$
jee-main 2019 Q85 Substitution to Evaluate a Definite Integral (Numerical Answer)
If $\int_0^{\pi/3} \frac{\tan\theta}{\sqrt{2k\sec\theta}}\,d\theta = 1 - \frac{1}{\sqrt{2}},\,(k > 0)$, then the value of $k$ is
(1) $\frac{1}{2}$
(2) 1
(3) 2
(4) 4
jee-main 2020 Q65 Substitution to Compute an Indefinite Integral with Initial Condition
The integral $\int \frac { d x } { ( x + 4 ) ^ { \frac { 8 } { 7 } } ( x - 3 ) ^ { \frac { 6 } { 7 } } }$ is equal to: (where $C$ is a constant of integration)
(1) $\left( \frac { x - 3 } { x + 4 } \right) ^ { \frac { 1 } { 7 } } + C$
(2) $\left( \frac { x - 3 } { x + 4 } \right) ^ { \frac { - 1 } { 7 } } + C$
(3) $\frac { 1 } { 2 } \left( \frac { x - 3 } { x + 4 } \right) ^ { \frac { 3 } { 7 } } + C$
(4) $- \frac { 1 } { 13 } \left( \frac { x - 3 } { x + 4 } \right) ^ { \frac { - 13 } { 7 } } + C$
jee-main 2020 Q66 Substitution to Compute an Indefinite Integral with Initial Condition
If $\int \left( e ^ { 2 x } + 2 e ^ { x } - e ^ { - x } - 1 \right) e ^ { \left( e ^ { x } + e ^ { - x } \right) } d x = g ( x ) e ^ { \left( e ^ { x } + e ^ { - x } \right) } + c$, where $c$ is a constant of integration, then $g ( 0 )$ is
(1) $e$
(2) $e ^ { 2 }$
(3) 1
(4) 2
jee-main 2021 Q74 Substitution to Transform Integral Form (Show Transformed Expression)
The value of the integral $\int \frac { \sin \theta \cdot \sin 2 \theta \left( \sin ^ { 6 } \theta + \sin ^ { 4 } \theta + \sin ^ { 2 } \theta \right) \sqrt { 2 \sin ^ { 4 } \theta + 3 \sin ^ { 2 } \theta + 6 } } { 1 - \cos 2 \theta } d \theta$ is (where $c$ is a constant of integration)
(1) $\frac { 1 } { 18 } \left[ 11 - 18 \sin ^ { 2 } \theta + 9 \sin ^ { 4 } \theta - 2 \sin ^ { 6 } \theta \right] ^ { \frac { 3 } { 2 } } + c$
(2) $\frac { 1 } { 18 } \left[ 9 - 2 \sin ^ { 6 } \theta - 3 \sin ^ { 4 } \theta - 6 \sin ^ { 2 } \theta \right] ^ { \frac { 3 } { 2 } } + c$
(3) $\frac { 1 } { 18 } \left[ 11 - 18 \cos ^ { 2 } \theta + 9 \cos ^ { 4 } \theta - 2 \cos ^ { 6 } \theta \right] ^ { \frac { 3 } { 2 } } + c$
(4) $\frac { 1 } { 18 } \left[ 9 - 2 \cos ^ { 6 } \theta - 3 \cos ^ { 4 } \theta - 6 \cos ^ { 2 } \theta \right] ^ { - \frac { 3 } { 2 } } + c$
jee-main 2021 Q74 Substitution to Evaluate a Definite Integral (Numerical Answer)
The value of the integral $\int _ { 0 } ^ { 1 } \frac { \sqrt { x } d x } { ( 1 + x ) ( 1 + 3 x ) ( 3 + x ) }$ is: (1) $\frac { \pi } { 4 } \left( 1 - \frac { \sqrt { 3 } } { 2 } \right)$ (2) $\frac { \pi } { 8 } \left( 1 - \frac { \sqrt { 3 } } { 6 } \right)$ (3) $\frac { \pi } { 8 } \left( 1 - \frac { \sqrt { 3 } } { 2 } \right)$ (4) $\frac { \pi } { 4 } \left( 1 - \frac { \sqrt { 3 } } { 6 } \right)$
jee-main 2021 Q75 Substitution to Transform Integral Form (Show Transformed Expression)
The integral $\int \frac { e ^ { 3 \log _ { e } 2 x } + 5 e ^ { 2 \log _ { e } 2 x } } { e ^ { 4 \log _ { e } x } + 5 e ^ { 3 \log _ { e } x } - 7 e ^ { 2 \log _ { e } x } } d x , x > 0$, is equal to (where $c$ is a constant of integration)
(1) $\log _ { \mathrm { e } } \left| x ^ { 2 } + 5 x - 7 \right| + \mathrm { c }$
(2) $4 \log _ { \mathrm { e } } \left| x ^ { 2 } + 5 x - 7 \right| + \mathrm { c }$
(3) $\frac { 1 } { 4 } \log _ { \mathrm { e } } \left| x ^ { 2 } + 5 x - 7 \right| + \mathrm { c }$
(4) $\log _ { e } \sqrt { x ^ { 2 } + 5 x - 7 } + c$