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.
 A: 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
$$I=\int \frac{dx}{(1-x)^2\sqrt{1-x^2}} dx = \frac{1}{4}\int (z^{-1/2}+z^{1/2}) dz=\frac{1}{2}z^{1/2}[1+\frac{z}{3}]=\frac{(2-x)}{3(1-x)} \sqrt{\frac{1+x}{1-x}}.$$
A: 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 $2(x-2)/(x-1)$.
A: 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,
$$ \begin{array}  {r c l} \displaystyle I := \int \dfrac1{ (1-x)^2\sqrt{1-x^2} } \, dx &=& \displaystyle \int \underbrace{\dfrac1{1-x}}_{=u} \cdot \underbrace{\dfrac1{ (1-x)\sqrt{1-x^2} } \, dx}_{=dv} \\ &=& \displaystyle \int u \, dv = uv - \int v \, du \\ 
&=& \dfrac{x+1}{(1-x)\sqrt{1-x^2}} - \displaystyle \int \dfrac{x+1}{\sqrt{1-x^2}} \cdot \dfrac1{(1-x)^2} \, dx \\ 
&=& \dfrac{x+1}{(1-x)\sqrt{1-x^2}} -      \displaystyle \int \dfrac{x}{(1-x)^2 \sqrt{1-x^2}}\, dx - \underbrace{\displaystyle \int \dfrac{1}{\sqrt{1-x^2}} \cdot \dfrac1{(1-x)^2} \, dx}_{=I} \\ 
2I &=& \dfrac{x+1}{(1-x)\sqrt{1-x^2}} + \displaystyle \int \dfrac{1-x - 1}{(1-x)^2 \sqrt{1-x^2}}\, dx \\
 &=& \dfrac{x+1}{(1-x)\sqrt{1-x^2}} + \displaystyle \int \dfrac{1}{(1-x)\sqrt{1-x^2}}\, dx - I  \\
3I &=& \dfrac{x+1}{(1-x)\sqrt{1-x^2}} + \dfrac{x+1}{\sqrt{1-x^2}} + C \\
I &=&  \dfrac{(x+1)(x-2)}{3(x-1)\sqrt{1-x^2}}  + C \\
\end{array}  $$
A: This kind of integral is classically calculated recursively:
Integrating by parts 
$\displaystyle\;d I_n=\smash[b]{\int\!\frac{\mathrm d x}{(1-x^2)^{\tfrac n2}}}\enspace (n \text{ odd})$, you obtain a relation between $I_n$ and $I_{n+2}$.
Here, setting $u=\dfrac 1{\sqrt{1-x^2}}$, $\;\mathrm dv=\mathrm dx$, whence $\;\mathrm du=\dfrac{x\mathrm d x}{(1-x^2)^{\frac 32}}$, $\;v = x$, you obtain
\begin{align}
\arcsin x=\int\!\frac{\mathrm dx}{\sqrt{1-x^2}}=\frac{x}{\sqrt{1-x^2}}+\int\!\dfrac{x^2\mathrm d x}{(1-x^2)^{\frac 32}}
\end{align}
Can you continue?
A: 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
$$I=\frac{1}{2} \int (v^2+1) dv= \frac{v}{2}[\frac{v^2}{3}+1]=\frac{\sqrt{2u-1}}{3} (u+1)=\frac{(2-x)}{3(1-x)} \sqrt{\frac{1+x}{1-x}}.$$
