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

$\frac{2}{π}[(2x-1)\sin^{-1}\sqrt{x}+\sqrt{x}\sqrt{1-x}]-x+C$

$I=\int(\frac{\sin^{-1}\sqrt{x}-\cos^{-1}\sqrt{x}}{\sin^{-1}\sqrt{x}+\cos^{-1}\sqrt{x}})dx=\int\frac{2}{π}(\sin^{-1}\sqrt{x}-\cos^{-1}\sqrt{x})=\int\frac{2}{π}(2\sin^{-1}\sqrt{x}-\frac{π}{2})dx$

$=\int(\frac{4}{π}\sin^{-1}\sqrt{x}-1)$. Put $x = \sin^2 t$ to get $dx = \sin 2t\, dt$

$⇒I=\int(\frac{4t}{π}-1)\sin 2t\,dt=\frac{4}{π}\int t\sin 2t\,dt-\int\sin 2t\,dt$

$=\frac{4}{π}[\frac{-t\cos\,2t}{2}-\int\frac{-\cos 2t}{2}dt]-(\frac{-\cos 2t}{2})=\frac{-2}{π}t\cos 2t+\frac{\sin 2t}{4}×\frac{4}{π}+\frac{\cos 2t}{2}+c$

$=\frac{-2}{π}\sin^{-1}\sqrt{x}(1-2x)+\frac{2\sqrt{x}\sqrt{1-x}}{4}×\frac{4}{π}+\frac{(1-2x)}{2}+c=\frac{2}{π}[\sin^{-1}\sqrt{x}(2x-1)+\sqrt{x}\sqrt{1-x}]-x+c$