## Calculus: Early Transcendentals 8th Edition

$arcsin~\frac{x-1}{x+1} = 2~arctan~\sqrt{x} -\frac{\pi}{2}$
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: $f(x) = arcsin~\frac{x-1}{x+1}-2~arctan~\sqrt{x} = -\frac{\pi}{2}$ $arcsin~\frac{x-1}{x+1} = 2~arctan~\sqrt{x} -\frac{\pi}{2}$