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 the Euler transformation $$z=\frac{1+x}{1-x}\Rightarrow x =\frac{z-1}{z+1}  \Rightarrow dx=\frac{2 dz}{(1+z)^2}$$ then
$$I=\int \frac{dx}{(1-x)^2\sqrt{1-x^2}} dx = \frac{1}{4}\int (z^{-1/2}+z^{1/2}) dz=\frac{1}{2}z^{1/2}[1+\frac{z}{3}]=\frac{(2-x)}{3(1-x)} \sqrt{\frac{1+x}{1-x}}.$$
