$I=\int {\sin }^{-1}\left(\frac{\sqrt{x}}{1+x}\right)dx$

$=\int \underset{I}{\underbrace{{\tan }^{-1}\left(\sqrt{x}\right)}}\underset{II}{\underbrace{\left(1\right)}}dx$

$=x{\tan }^{-1}\sqrt{x}-\int \frac{1}{1+x}.\frac{1}{2\sqrt{x}}.xdx+C$

$=x{\tan }^{-1}\sqrt{x}-\frac{1}{2}\int \frac{t.2t.dt}{1+t^2}+C$ $($putting $x=t^2\Rightarrow dx=2tdt)$

$=x{\tan }^{-1}\sqrt{x}-\int \frac{t^2}{1+t^2}dt+C$

$=x{\tan }^{-1}\sqrt{x}-t+{\tan }^{-1}t+C$

$=x{\tan }^{-1}\sqrt{x}-\sqrt{x}+{\tan }^{-1}\sqrt{x}+C$

$=\left(x+1\right){\tan }^{-1}\sqrt{x}-\sqrt{x}+C$

Comparing with $A\left(x\right){\tan }^{-1}\left(\sqrt{x}\right)+B\left(x\right)+C$, we get

$\left(\mathrm{A}\left(\mathrm{x}\right), \mathrm{B}\left(\mathrm{x}\right)\right)=\left(x+1, -\sqrt{x}\right)$