Question

Integrate with respect to $x$:
$\frac{x^2{\tan }^{-1}x}{1+x^2}$

Answer 1

Consider the given integral.

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

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

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

Therefore,

$I=\int \left({\tan }^{2}t\right)tdt$

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

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

$I=t\tan t-\int 1\tan tdt-\frac{t^2}{2}$

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

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

On putting the value of $t$, we get

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

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

Hence, this is the answer.