Finding the integral $\int \frac{dx}{(1-x)^2\sqrt{1-x^2}}$ One may take $x= \cos t$ and get
$$I=\int \frac{dx}{(1-x)^2\sqrt{1-x^2}}= -\frac{1}{4}\int \csc^4(t/2)~ dt=-\frac{1}{4} \int [\csc^2(t/2) +\csc^2(t/2) \cot^2(t/2)]~ dt.$$
$$\Rightarrow I=\frac{1}{2} \left [\cot (t/2)] +\frac{1}{3}\cot^3(t/2)\right]=\frac{(2-x)}{3(1-x)} \sqrt{\frac{1+x}{1-x}}.$$
The question is: How to obtain this integral by other methods not using the trigonometric substitution.

Because we have the radical $\sqrt{1-x^2}=\sqrt{(1-x)(1+x)}$ (and because we know the anser), we may try the Euler substitution $$
t=\sqrt{\frac {1+x}{1-x}}\ .
$$
In 
https://en.wikipedia.org/wiki/Euler_substitution
it is the third Euler substitution:
$$
\sqrt{(1+x)(1-x)}=t(1-x)\ .
$$
Then $t^2=(1+x)/(1-x)$, $t^2(1-x)=1+x$, $x(t^2+1)=t^2-1$, 
$$
x =\frac{t^2-1}{t^2+1}=
1-\frac2{t^2+1}\ ,
$$
so we can express $x$ rationally in terms of $t$, so also $t(1-x)$, which is the radical.
Putting all together:
$$
\begin{aligned}
\int\frac 1{(1-x)^2\sqrt{1-x^2}}\; dx
&=
\int\frac 1{\frac4{(t^2+1)^2}\cdot \frac {2|t|}{t^2+1}}\;\frac{4t}{(t^2+1)^2}\; dt
\\
&\qquad\text{(and i will assume $t>0$...)}
\\
&=\int \frac 12(t^2+1)\; dt
\\
&=\frac 16t^3+\frac 12 t+\text{locally constant function .}
\\
&=\frac 16\cdot t\cdot(t^2+3)+\text{locally constant function .}
\end{aligned}
$$
This matches, since $t^2+3$ is $2(x-2)/(x-1)$.
