$\int \frac{{\sin }^{-1}x-{\cos }^{-1}x}{{\sin }^{-1}x+{\cos }^{-1}x}dx=$

$\int \frac{{\sin }^{-1}x-{\cos }^{-1}x}{{\sin }^{-1}x+{\cos }^{-1}x}dx$

$=\int \frac{{\sin }^{-1}x-\left(\frac{\pi }{2}-{\sin }^{-1}x\right)}{\pi /2}dx \left(\because {\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}\right)$

$=\frac{2}{\pi }\int \left(2{\sin }^{-1}x-\frac{\pi }{2}\right)dx=\frac{4}{\pi }\int {\sin }^{-1}x1.dx-\int 1dx$

$=\frac{4}{\pi }\left[{\sin }^{-1}x.x-\int \frac{{\displaystyle 1}}{{\displaystyle \sqrt{1-x^2}}}xdx\right]-x+c$

$=\frac{4}{\pi }\left[x{\sin }^{-1}x+\frac{{\displaystyle 1}}{{\displaystyle 2}}\int \frac{{\displaystyle -2x}}{{\displaystyle \sqrt{1-x^2}}}dx\right]-x+c$

$=\frac{4}{\pi }\left[x{\sin }^{-1}x+\frac{{\displaystyle 1}}{{\displaystyle 2}}2\sqrt{1-x^2}\right]-x+c$

$=\frac{4}{\pi }\left[x{\sin }^{-1}x+\sqrt{1-x^2}\right]-x+c$