If $I_1={\displaystyle \int Si{n}^{-1}\left(\frac{2x}{1+x^2}\right)dx,I_2={\displaystyle \int Co{s}^{-1}\left(\frac{1-x^2}{1+x^2}\right)dx,I_3={\displaystyle \int Ta{n}^{-1}\left(\frac{2x}{1-x^2}\right)dx}}}$ , then $I_1+I_2-I_3=$

$={\displaystyle \int 2Ta{n}^{-1}x}dx$

$I_2={\displaystyle \int Co{s}^{-1}\left(\frac{1-x^2}{1+x^2}\right)dx}$

$={\displaystyle \int 2Ta{n}^{-1}x}$

$\therefore I_1+I_2-I_3={\displaystyle \int 2Ta{n}^{-1}xdx}$

$=2\left[Ta{n}^{-1}x.x-\frac{1}{2}{\displaystyle \int \frac{1}{1+x^2}.2xdx}\right]$

$=2\left[x.Ta{n}^{-1}x-\frac{1}{2}\log \left(1+x^2\right)\right]c$

$=2xTa{n}^{-1}x-\log \left|1+x^2\right|+c$