Question

Find: $\int x{\tan }^{-1}xdx$

Answer 1

$\int f\left(x\right)\dot{g\left(x\right)}=f\left(x\right)g\left(x\right)-\int \dot{f\left(x\right)}g\left(x\right)$

$f\left(x\right)={\tan }^{-1}x$ and $\dot{g\left(x\right)}=xg\left(x\right)=\frac{x^2}{2}$

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

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