<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] Hints:
1. Substitute, as you did, $u = 1-x$, $\text{d}u = -\text{d}x$
2. you get $\int\frac{1}{(u-1)\sqrt{u}}\ \text{d}u$
3. Substitute now $y = \sqrt{u}$, $\text{d}y = \frac{1}{2\sqrt{u}}\ \text{d}u$
4. Integral is now $-2\int\frac{1}{1 - y^2}\ \text{d}y$
5. Solution of that integration is $-2\ \text{arctanh}(y) + C$
Now you can substitute back to $x$ and get your answer
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