Answer
$arcsin~\frac{x-1}{x+1} = 2~arctan~\sqrt{x} -\frac{\pi}{2}$
Work Step by Step
Let $f(x) = arcsin~\frac{x-1}{x+1}-2~arctan~\sqrt{x}$
We can find $f'(x)$:
$f'(x) = \frac{[(x+1)-(x-1)]}{(x+1)^2\sqrt{1-(\frac{x-1}{x+1})^2}}-\frac{2}{2~(1+x)~\sqrt{x}}$
$f'(x) = \frac{2}{(x+1)^2\sqrt{\frac{(x+1)^2}{(x+1)^2}-(\frac{(x-1)^2}{(x+1)^2}}}-\frac{1}{(1+x)~\sqrt{x}}$
$f'(x) = \frac{2}{\frac{(x+1)^2}{x+1}\sqrt{(x+1)^2-(x-1)^2}}-\frac{1}{(1+x)~\sqrt{x}}$
$f'(x) = \frac{2}{(x+1)~\sqrt{4x}}-\frac{1}{(1+x)~\sqrt{x}}$
$f'(x) = \frac{1}{(x+1)~\sqrt{x}}-\frac{1}{(1+x)~\sqrt{x}}$
$f'(x) = 0$
Therefore, $f(x) = C$, where $C$ is some constant.
Consider $f(0)$:
$f(x) = arcsin~\frac{x-1}{x+1}-2~arctan~\sqrt{x}$
$f(0) = arcsin~\frac{0-1}{0+1}-2~arctan~\sqrt{0}$
$f(0) = arcsin~(-1)-2~arctan~0$
$f(0) = -\frac{\pi}{2}-0$
$f(0) = -\frac{\pi}{2}$
Thus $C = -\frac{\pi}{2}$
Then:
$f(x) = arcsin~\frac{x-1}{x+1}-2~arctan~\sqrt{x} = -\frac{\pi}{2}$
$arcsin~\frac{x-1}{x+1} = 2~arctan~\sqrt{x} -\frac{\pi}{2}$