The integral $\int x{\cos }^{-1}\left(\frac{{\displaystyle 1-x^2}}{{\displaystyle 1+x^2}}\right)dx(x>0)$ is equal to 

$\int x{\cos }^{-1}\left(\frac{{\displaystyle 1-x^2}}{{\displaystyle 1+x^2}}\right)dx$

$Put x=\tan \theta \Rightarrow dx=se{c}^2\theta d\theta$

$$\begin{gathered} =\int \tan \theta {\cos }^{-1}\left(\frac{1-\tan \theta }{1+{\tan }^2\theta }\right)se{c}^2\theta d\theta \\ =\int \tan \theta {\cos }^{-1}\left(\cos 2\theta \right)se{c}^2\theta d\theta \\ =\int 2\theta .\left(\tan \theta . se{c}^2\theta \right)d\theta \\ \left(\because by parts\right) \\ =\left(2\theta \right)\frac{{\tan }^2\theta }{2}-\int \left(2\right).\frac{{\tan }^2\theta }{2}d\theta \\ \left(\therefore \int \tan \theta se{c}^2\theta d\theta =\int \tan \theta d\tan \theta =\frac{{\displaystyle {\left(\tan \theta \right)}^2}}{2}+C\right) \\ =\theta .{\tan }^2\theta -\int \left(se{c}^2\theta -1\right)d\theta \\ =\theta {\tan }^2\theta -\left(\tan \theta -\theta \right)+C \\ =\left({\tan }^{-1}x\right)x^2-x+{\tan }^{-1}x+C \\ =-x+{\tan }^{-1}x\left(x^2+1\right)+C \end{gathered}$$