<s>[INST] 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
 [/INST] A yet different approach:
\begin{align}
I&=\int \frac{dx}{x\sqrt{1-x}}\\
&=\int \frac{-2t^{-2}+2t^{-3}}{(2t^{-1}-t^{-2})\sqrt{1-2t^{-1}+t^{-2}}}dt \mbox{, where $x=2t^{-1}-t^{-2}$}\\ 
&=\int \frac{-2t^{-2}(1-t^{-1})}{(2t^{-1}-t^{-2})(1-t^{-1})}dt \\
&=-2\int \frac{1}{2t-1}dt \\
&=-\ln(2t-1)\\
&=\ln\frac{x}{2+2\sqrt{1-x}-x} \mbox{, where $t=\frac{1+\sqrt{1-x}}{x}$}\\
&=\ln\frac{(1-\sqrt{1-x})(1+\sqrt{1-x})}{(1+\sqrt{1-x})^2}\\
&=\ln\frac{1-\sqrt{1-x}}{1+\sqrt{1-x}}\\
\end{align}
</s>