Integrate $\sqrt \frac{x}{1-x} $ What would be the best method to integrate $\sqrt \frac{x}{1-x} $ ?
I tried the substitution $1-x=t $ that makes the inner fraction $(1/t-1)$.But after that not getting any simple method.Please help me out...
 A: Hint. One may perform the change of variable
$$
u=\sqrt \frac{x}{1-x},\quad x= \frac{u^2}{u^2+1}, \quad dx= \frac{2u}{(u^2+1)^2}du,
$$ giving
$$
\int\sqrt \frac{x}{1-x}\:dx=2\int \frac{u^2}{(u^2+1)^2}du=-\frac{u}{1+u^2}+\arctan u+C
$$ or

$$
\int\sqrt \frac{x}{1-x}\:dx=-\sqrt{x(1-x)}+\arctan \left( \sqrt \frac{x}{1-x}\right)+C.
$$

A: Running with Andre's excellent suggestion to use $x=\sin^2 \theta$ (notice the immediate simplification of the denominator!) we get the form
$$\int \sqrt{\frac{x}{1-x}} dx = \int \frac{d\theta}{\sin(2\theta)}\sqrt{\frac{\sin^2\theta}{\cos^2\theta}} =\int\tan\theta\sin(2\theta)d\theta$$
Note that above we require that $x\in [0,1)$ and also that $0\leq \theta< \frac{\pi}{2}$ and thus we drop the absolute value (the domain restriction on $\theta$ allows the substitution to be a function, satisfying our needs)
$$\int\tan\theta\sin(2\theta)d\theta = \theta-\frac{\sin(2\theta)}{2}$$

$$= \;\arcsin(\sqrt x)-\sqrt{1-x}\sqrt x$$

A: Hint. Let $u=\sqrt {1-x} $, then $u^2 = 1-x $ ,$x = 1-u^2$ ,$dx =-2udu $
The integral will be $\frac{\sqrt{1-u^2}}{u}\times-2udu = -2\sqrt{1-u^2}du$.
Try use trigonometric function to substitute $u$.
Easy to show that: $-2\int \sqrt{1-u^2}du = -u \sqrt{1-u^2}- \arcsin(u) + C = -\sqrt{x-x^2} -\arcsin(\sqrt{1-x})+C.$
A: Hint
Substitute $$1-x= t^2$$ You will get something like $$\int \sqrt {(1-t^2)} dx$$
Then substitute $ t= sin(\theta)$
