How to evaluate $\int \frac{dx}{(1-x)\sqrt{1-x^2}}$ without a trig substitution/parts? I'd like to find 
$$\int \frac{dx}{(1-x)\sqrt{1-x^2}}$$
but without using a trig substitution or integration by parts. I can already see that $x=\sin \theta$ works it out quite nicely, but I was hoping for an alternative solution. I tried usual suspects like $u=\frac{1}{x}$ but I didn't get very far with them. What could I perhaps try instead?
 A: Let
$$u=\frac{1}{1-x}$$
Then,
$$du=\frac{1}{(1-x)^2}dx$$
and
$$u-ux=1$$
$$x=1-\frac1u$$
Thus,
$$\int \frac{dx}{(1-x)\sqrt{1-x^2}}=\int \frac{(1-x)dx}{(1-x)^2\sqrt{1-x^2}}\int\frac{du}{u\sqrt{(1-(1-\frac1u)^2)}}=\int\frac{du}{u\sqrt{(1-1-\frac1{u^2}+\frac2u)}}=\int\frac{du}{u\sqrt{(-\frac1{u^2}+\frac2u)}}=\int\frac{1}{\sqrt{2u-1}}du$$
which can be easily solved.
A: Here is another approach which does not make use of a trigonometric substitution.
\begin{align*}
\int \frac{dx}{(1 - x)\sqrt{1 - x^2}} &= \int \frac{dx}{(1 - x)\sqrt{(1 - x)(1 + x)}}\\
&= \int \frac{1}{(1 - x)\sqrt{(1 - x)(1 + x)}} \cdot \frac{\sqrt{1 - x}}{\sqrt{1 - x}} \, dx\\
&= \int \sqrt{\frac{1 - x}{1 + x}} \frac{dx}{(1 - x)^2}
\end{align*}
Now using the self-similar substitution of $u = \dfrac{1 - x}{1 + x}$, we see that $x = \dfrac{1 - u}{1 + u}$, giving $dx = -\dfrac{2}{(1 + u)^2} \, du$.
Noting that $1 - x = \dfrac{2u}{1 + u}$, the integral becomes
\begin{align*}
\int \frac{dx}{(1 - x)\sqrt{1 - x^2}} &= \int \sqrt{u} \cdot \left (\frac{1 + u}{2u} \right )^2 \cdot -\frac{2}{(1 + u)^2} \, dx\\
&= -\frac{1}{2} \int u^{-3/2} \, du\\
&= \frac{1}{\sqrt{u}} + {\cal{C}}\\
&= \sqrt{\frac{1 + x}{1 - x}} + \cal{C}.
\end{align*} 
