Why is $\int \frac{1}{\sqrt{y}\sqrt{1 - y}} dy = \frac{2\sqrt{y - 1}\sqrt{y} \log(\sqrt{y - 1} + \sqrt{y})}{\sqrt{(-(y - 1) y)}} $? Fairly self-explanatory question title. Why is
$$\int \frac{1}{\sqrt{y}\sqrt{1 - y}} dy = \frac{2\sqrt{y - 1}\sqrt{y} \log(\sqrt{y - 1} + \sqrt{y})}{\sqrt{-(y - 1)}\sqrt{y}}\ ? $$  
I'm assuming you have to use substition, but I'm not sure how.
edit: $$ y \in (0,1) $$

Is the answer correct? Notice $y\in (0,1)$, as both $y> 0$, $1-y>0$. But why answer has $\sqrt{y-1}$?
Let $y=\sin ^2x$, $x\in (0,\frac{\pi}{2})$
$$\int \frac{1}{\sqrt{y}\sqrt{1 - y}} dy = \int \frac{2\sin x \cos x}{\sin x\cos x} dx =2x +C=2\arcsin \sqrt{y}+C$$
