Question

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

• A. $(x-1,-\sqrt{x})$
• B. $(x+1,\sqrt{x})$
• C. $(x+1,-\sqrt{x})$
• D. $(x-1,\sqrt{x})$

Answer 1

The correct option is C $(x+1,-\sqrt{x})$

$\int {\sin }^{-1}\sqrt{\frac{x}{1+x}}dx$
$=\int \underset{I}{{\tan }^{-1}}\sqrt{x}\cdot \underset{II}{1}dx$
$=\left({\tan }^{-1}\sqrt{x}\right)\cdot x-\int \frac{x}{1+x}\cdot \frac{1}{2\sqrt{x}}dx$
Put $x=t^2\Rightarrow dx=2tdt$
$=x{\tan }^{-1}\sqrt{x}-\int \frac{\left(t^2\right)\left(2t\right)dt}{(1+t^2)\left(2t\right)}$
$=x{\tan }^{-1}\sqrt{x}-t+{\tan }^{-1}t+C$
$=x{\tan }^{-1}\sqrt{x}-\sqrt{x}+{\tan }^{-1}\sqrt{x}+C$
$=(x+1){\tan }^{-1}\sqrt{x}-\sqrt{x}+C$
$\therefore A\left(x\right)=x+1,B\left(x\right)=-\sqrt{x}$