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.

Let us first find evaluate the indefinite integral $ \displaystyle \int \dfrac1{ (1-x)\sqrt{1-x^2} } \, dx $.
$$\begin{array} {r c l} \displaystyle \int \dfrac1{ (1-x)\sqrt{1-x^2} } \, dx &=& \displaystyle  \int \dfrac{1+x}{ (1-x^2)\sqrt{1-x^2} } \, dx  \\ &=& \displaystyle  \int \dfrac{1}{ (1-x^2)\sqrt{1-x^2} } \, dx + \displaystyle \int \dfrac{x}{ (1-x^2)\sqrt{1-x^2} } \, dx \end{array} $$
Using the substitution $y = \sqrt{1-x^2}$ for both these integrals, we can find that they evaluate to $ \frac x{\sqrt{1-x^2}} $ and $ \frac1{\sqrt{1-x^2}} $, respectively. Thus, $$ \displaystyle \int \dfrac1{ (1-x)\sqrt{1-x^2} } \, dx   = \frac{x+1}{\sqrt{1-x^2}} + C .$$
Now we integrate by parts,
$$ \begin{array}  {r c l} \displaystyle I := \int \dfrac1{ (1-x)^2\sqrt{1-x^2} } \, dx &=& \displaystyle \int \underbrace{\dfrac1{1-x}}_{=u} \cdot \underbrace{\dfrac1{ (1-x)\sqrt{1-x^2} } \, dx}_{=dv} \\ &=& \displaystyle \int u \, dv = uv - \int v \, du \\ 
&=& \dfrac{x+1}{(1-x)\sqrt{1-x^2}} - \displaystyle \int \dfrac{x+1}{\sqrt{1-x^2}} \cdot \dfrac1{(1-x)^2} \, dx \\ 
&=& \dfrac{x+1}{(1-x)\sqrt{1-x^2}} -      \displaystyle \int \dfrac{x}{(1-x)^2 \sqrt{1-x^2}}\, dx - \underbrace{\displaystyle \int \dfrac{1}{\sqrt{1-x^2}} \cdot \dfrac1{(1-x)^2} \, dx}_{=I} \\ 
2I &=& \dfrac{x+1}{(1-x)\sqrt{1-x^2}} + \displaystyle \int \dfrac{1-x - 1}{(1-x)^2 \sqrt{1-x^2}}\, dx \\
 &=& \dfrac{x+1}{(1-x)\sqrt{1-x^2}} + \displaystyle \int \dfrac{1}{(1-x)\sqrt{1-x^2}}\, dx - I  \\
3I &=& \dfrac{x+1}{(1-x)\sqrt{1-x^2}} + \dfrac{x+1}{\sqrt{1-x^2}} + C \\
I &=&  \dfrac{(x+1)(x-2)}{3(x-1)\sqrt{1-x^2}}  + C \\
\end{array}  $$
