integrate $\int \frac{dx}{x\sqrt{1-x}}$ 
$$\int \frac{dx}{x\sqrt{1-x}}$$

$$\int \frac{dx}{x\sqrt{1-x}}$$
$u=1-x$
$du=-dx$
$$-\int \frac{du}{(1-u)\sqrt{u}}$$
$a(1-u)+b\sqrt{u}=1\Rightarrow a-au+b\sqrt{u}=1$
$a=1\Rightarrow b\sqrt{u}-u=0\Rightarrow b=\sqrt{u}$
$$\int \frac{\sqrt{u}}{1-u}du-\int \frac{du}{\sqrt{u}}=\int \frac{\sqrt{u}}{1-u}du-2{\sqrt{u}}$$
$$\int \frac{\sqrt{u}}{1-u}du=\int \frac{1+\sqrt{u}-1}{(1+\sqrt{u})(1-\sqrt{u})}du=\int \frac{1+\sqrt{u}}{(1+\sqrt{u})(1-\sqrt{u})}du-\int \frac{du}{(1+\sqrt{u})(1-\sqrt{u})}du=\int \frac{1}{(1-\sqrt{u})}du-\int \frac{du}{(1+\sqrt{u})(1-\sqrt{u})}du$$
$$\int \frac{1}{(1-\sqrt{u})}du$$
How do I continue from here? it seems that I have made it harder

HINT:
Let $ x = u^2, \, dx= 2 u du ; \,  \int \dfrac{2  du }{u\sqrt{1-u^2} }$
again let  $$ u = \sin(t) ; \ du = \cos (t) dt  ...$$
leading to 
$$  =2 log \dfrac{u}{1+\sqrt{1-u^2}} = 2 log \dfrac{\sqrt x}{1+\sqrt{1-x} }$$
