Mathematics

# $\int_{}^{} {\cfrac{{{x^2}{{\tan }^{ - 1}}{x^3}}}{{1 + {x^6}}}} dx$

##### SOLUTION
We have,
$I=\int_{}^{} {\cfrac{{{x^2}{{\tan }^{ - 1}}{x^3}}}{{1 + {x^6}}}} dx$

Let
$t=\tan^{-1}x^3$

$\dfrac{dt}{dx}=\dfrac{1}{1+(x^3)^2}\times 3x^2$

$\dfrac{dt}{3}=\dfrac{x^2}{1+x^6}dx$

Therefore,
$I=\dfrac{1}{3}\int t\ dt$

$I=\dfrac{1}{3}\dfrac{t^2}{2}+C$

On putting the value of $t$, we get
$I=\dfrac{1}{6}(\tan^{-1}x^3)^2+C$

Its FREE, you're just one step away

Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

Q1 Subjective Medium
for any natural number m, evaluate
$\int {({x^{3m}} + {x^{2m}} + {x^m})} \,\,\,{(2{x^{2m}} + 3{x^m} + 6)^\cfrac{1}{m}}$ $dx,x>0$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Evaluate the following definite integral:

$\displaystyle \int_{e}^{e^2} \left\{\dfrac {1}{\log x} -\dfrac {1}{(\log x)^2}\right\} dx$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Medium
If $\displaystyle \int { \frac { { e }^{ x }-1 }{ { e }^{ x }+1 } dx } =f\left( x \right) +C$, then $f\left( x \right)$ is equal to
• A. $\log { \left( { e }^{ 2x }-1 \right) }$
• B. $2\log { \left( { e }^{ x }+1 \right) } -x$
• C. $\log { \left( { e }^{ x }-1 \right) }$
• D. $2\log { \left( { e }^{ x }+1 \right) }$

1 Verified Answer | Published on 17th 09, 2020

Q4 Single Correct Medium
The value of $\displaystyle \int_0^1 \sqrt{x} \cdot e^{\sqrt{x}} dx$ is equal to
• A. $\dfrac {(e-2)}{2}$
• B. $2e-1$
• C. $2 (e-1)$
• D. $\dfrac {e-1}{2}$
• E. $2 ( e-2)$

Evaluate $\int \dfrac{e^x-e^{-x}}{e^x+e^{-x}}dx$