Question

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

Answer 1

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

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

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

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

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

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

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

$I=t.\tan t-\log \sec t-\frac{t^2}{2}+C$

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

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