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 use $$1-x=\frac{1}{u} \Rightarrow x=1-\frac{1}{u} \Rightarrow dx=\frac{du}{u^2}.$$ Then
$$I=\int \frac{dx}{(1-x)^2\sqrt{1-x^2}}= \int \frac{u du}{\sqrt{2u-1}}.$$
Next use $$2u-1=v^2 \Rightarrow du =v dv.$$ Then
$$I=\frac{1}{2} \int (v^2+1) dv= \frac{v}{2}[\frac{v^2}{3}+1]=\frac{\sqrt{2u-1}}{3} (u+1)=\frac{(2-x)}{3(1-x)} \sqrt{\frac{1+x}{1-x}}.$$
