$If \int {\sin }^{-1}\left(\sqrt{\frac{x}{1+x}}\right)\mathrm{dx}=\mathrm{A}\left(\mathrm{x}\right){\tan }^{-1}\left(\sqrt{x}\right)+\mathrm{B}\left(\mathrm{x}\right)+c, \text{ where c is a constant of integration, then the ordered pair }$(A(x), B(x))can be :

$\text{ consider the integral }\int {\sin }^{-1}\sqrt{\frac{x}{1+x}}dx$

$\begin{array}{l}\text{ Use the triangle, }\int {\sin }^{-1}\sqrt{\frac{x}{1+x}}dx=\int {Tan}^{-1}\sqrt{x}dx\\ \text{ Problem becomes }\int {Tan}^{-1}\sqrt{x}dx\end{array}$

By using by parts,

$\begin{array}{c}\int Ta{n}^{-1}\sqrt{x}dx=x.{\tan }^{-1}\sqrt{x}-\int \frac{x}{1+x}.\frac{1}{2\sqrt{x}}dx\\ =x.{\tan }^{-1}\sqrt{x}-\int \left(\frac{1}{2\sqrt{x}}-\frac{1}{\left(1+{\left(\sqrt{x}\right)}^2\right)}.\frac{1}{2\sqrt{x}}\right)dx\\ =x.{\tan }^{-1}\sqrt{x}-\sqrt{x}+Ta{n}^{-1}\sqrt{x}+c\\ =\left(x+1\right)Ta{n}^{-1}\sqrt{x}-\sqrt{x}+c\end{array}$

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