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)}} $?

Question

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: $$ y \in (0,1) $$

Answer 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 $y=\sin ^2x$, $x\in (0,\frac{\pi}{2})$ $$\int \frac{1}{\sqrt{y}\sqrt{1 - y}} dy = \int \frac{2\sin x \cos x}{\sin x\cos x} dx =2x +C=2\arcsin \sqrt{y}+C$$

Answer 2

Hint:

As $$4y(1-y)=1-(2y-1)^2$$

Set $2y-1=\sin t$

Answer 3

Another possibility is to write it as $$\int\frac{\sqrt{y}}{y\sqrt{1-y}}dy$$and substitute $$u=\sqrt{\frac{y}{1-y}}$$ this avoids trigonometry.