Question

Evaluate $\int \frac{x^2{\tan }^{-1}x}{1+x^2}dx$

Answer 1

Consider the given integral.

$I=\int \frac{x^2\left({\tan }^{-1}x\right)}{1+x^2}dx$

Put

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

$dt=\frac{1}{1+x^2}dx$

Therefore,

$I=\int t.{\tan }^{2}tdt$

$I=\int t.({\sec }^{2}t-1)dt$

$I=\int (t{\sec }^{2}t-t)dt$

$I=\int t{\sec }^{2}tdt-\int tdt$

$I=t\tan t-\int 1.tantdt-\frac{t^2}{2}+C$

$I=t\tan t-[-\log \left|\cos t\right|]-\frac{t^2}{2}+C$

$I=t\tan t+\log \left|\cos t\right|-\frac{t^2}{2}+C$

Put the value of $t$, we get

$I={\tan }^{-1}x\tan \left({\tan }^{-1}x\right)+\log \left|\cos \left({\tan }^{-1}x\right)\right|-\frac{{\left({\tan }^{-1}x\right)}^2}{2}+C$

$I=x{\tan }^{-1}x+\log \left|\cos \left({\tan }^{-1}x\right)\right|-\frac{{\left({\tan }^{-1}x\right)}^2}{2}+C$

Hence, this is the required value of integral.