<s>[INST] Evaluate $\int_0^1 \frac{\mathrm{d}x}{\sqrt{1-x^2}}\frac{x }{1-k^2x^2}\log\left(\frac{1-x}{1+x}\right)$ I am trying to evaluate the following integral
$$I(k) = \int_0^1 \frac{\mathrm{d}x}{\sqrt{1-x^2}}\frac{x }{1-k^2x^2}\log\left(\frac{1-x}{1+x}\right)$$
with $0< k < 1$.
My attempt
By performing the substitution 
$$y=\frac{1-x}{1+x} \Longleftrightarrow x=\frac{1-y}{1+y}$$
we have
$$ I(k) = \int_0^1 \mathrm{d}y \frac{\log y}{\sqrt{y}}\frac{y-1}{(k^2-1)y^2-2y(k^2+1)+(k^2-1)}$$
Now we can decompose
$$\frac{y-1}{(k^2-1)y^2-2y(k^2+1)+(k^2-1)}= \frac{1}{2k(1+k)}\frac{1}{y+a}- \frac{1}{2k(1-k)}\frac{1}{y+a^{-1}}$$
with $a=\frac{1-k}{1+k}$ runs from $0$ to $1$.
Therefore we can write
$$I(k) = \frac{1}{2k(1+k)}\color{blue}{\int_0^1 \mathrm{d}y \frac{\log y}{\sqrt{y}}\frac{1}{y+a}} - \frac{1}{2k(1-k)}\color{red}{\int_0^1 \mathrm{d}y \frac{\log y}{\sqrt{y}}\frac{1}{y+a^{-1}}}$$
So we have only to evaluate $\color{blue}{I_1}$ and $\color{red}{I_2}$. 
Now we consider the following integral
$$ J(\sigma) = \int_0^1 \mathrm{d}x \frac{\log x}{\sqrt{x}}\frac{1}{x-\sigma^2}$$
so that $\color{blue}{I_1}=J(i\sqrt{a})$ and $\color{red}{I_2}=J(i/\sqrt{a})$.
By considering the map $x\mapsto x^2$ we can write
$$ J(\sigma)=4\int_0^1\mathrm{d}x \frac{\log x}{x^2-\sigma^2}= \frac{2}{\sigma}\left[\color{green}{\int_0^1 \frac{\log x}{x-\sigma}}-\color{green}{\int_0^1 \frac{\log x}{x+\sigma}}\right]$$
The problem then reduces to evaluate the $\color{green}{\text{green}}$ integrals. At this point I'm stuck. I think that it needs to be solved by using polylogarithms, but I don't really know how to use these functions.
Mathematica 11.0 says
$$J(\sigma)=4 \left(\frac{\Phi \left(\frac{1}{\sigma ^2},2,\frac{3}{2}\right)}{4 \sigma ^4}+\frac{1}{\sigma ^2}\right)$$
where $\Phi$ is the  Lerch transcendent. I don't know if this result is true (numerical integration is somewhat problematic). However, if it is true, I don't know what to do next.

Any hint on how to proceed with the evaluation?

Thanks in advance!
 [/INST] I have completed the computation of ASM.
For $0<k<1$,
$\displaystyle I(k) = \int_0^1 \frac{\mathrm{d}x}{\sqrt{1-x^2}}\frac{x }{1-k^2x^2}\log\left(\frac{1-x}{1+x}\right)\,dx$
Let $\rho=\dfrac{1-k}{1+k}$
Observe that, since $0<k<1$ then $0<\rho<1$.
Perform the change of variable $y=\dfrac{1-x}{1+x}$,
$\begin{align}I(k)&=\int_0^1 \frac{(y-1)\ln y}{\sqrt{y}\big(\left(k-1\right)y-k-1\big)\big(\left(k+1\right)y-k+1\big)}\,dy\\
&=\frac{1}{(k^2-1)}\int_0^1 \frac{(y-1)\ln y}{\sqrt{y}\left(y+\frac{1}{\rho}\right)\left(y+\rho\right)}\,dy
\end{align}$
Perform the change of variable $\displaystyle t=\sqrt{y}$,
$\begin{align}I(k)&=\frac{4}{(k^2-1)}\int_0^1 \frac{(t^2-1)\ln t}{\left(t^2+\frac{1}{\rho}\right)\left(t^2+\rho\right)}\,dt\\
&=\frac{4\rho}{(\rho-1)(k^2-1)}\left(\int_0^1 \frac{\ln t}{t^2+\rho}\,dt-\int_0^1 \frac{\ln t}{\rho t^2+1}\,dt\right)\\
\end{align}$
In the latter integral perform the change of variable $\displaystyle u=\frac{1}{t}$,
$\begin{align}I(k)&=\frac{4\rho}{(\rho-1)(k^2-1)}\left(\int_0^1 \frac{\ln t}{t^2+\rho}\,dt+\int_1^{\infty} \frac{\ln u}{u^2+\rho}\,du\right)\\
&=\frac{4\rho}{(\rho-1)(k^2-1)}\int_0^{\infty} \frac{\ln t}{t^2+\rho}\,dt\\
\end{align}$
Since $\displaystyle \rho>0$,
$\begin{align}I(k)&=\frac{4}{(\rho-1)(k^2-1)}\int_0^{\infty} \frac{\ln t}{\left(\frac{t}{\sqrt{\rho}}\right)^2+1}\,dt\end{align}$
Perform the change of variable $v=\dfrac{t}{\sqrt{\rho}}$,
$\begin{align}I(k)&=\frac{4\sqrt{\rho}}{(\rho-1)(k^2-1)}\int_0^{\infty} \frac{\ln\left( v\sqrt{\rho}\right)}{v^2+1}\,dv\\
&=\frac{2\sqrt{\rho}\ln\left(\rho\right)}{(\rho-1)(k^2-1)}\int_0^{\infty} \frac{1}{v^2+1}\,dv+\frac{4\sqrt{\rho}}{(\rho-1)(k^2-1)}\int_0^{\infty} \frac{\ln\left( v\right)}{v^2+1}\,dv\\
&=\frac{2\sqrt{\rho}\ln\left(\rho\right)}{(\rho-1)(k^2-1)}\Big[\arctan v\Big]_0^{\infty}+\frac{4\sqrt{\rho}}{(\rho-1)(k^2-1)}\int_0^{\infty} \frac{\ln\left( v\right)}{v^2+1}\,dv\\
&=\frac{\pi\sqrt{\rho}\ln\left(\rho\right)}{(\rho-1)(k^2-1)}+\frac{4\sqrt{\rho}}{(\rho-1)(k^2-1)}\int_0^{\infty} \frac{\ln\left( v\right)}{v^2+1}\,dv\\
\end{align}$
But,
Consider the integral,
$\displaystyle J=\int_0^{\infty} \frac{\ln\left( v\right)}{v^2+1}\,dv$
Perform the change of variable $w=\dfrac{1}{v}$,
$\displaystyle J=-J$.
Therefore $\displaystyle J=0$,
and,
$\begin{align}J(k)&=\frac{\pi\sqrt{\rho}\ln\left(\rho\right)}{(\rho-1)(k^2-1)}\\
&=\frac{\pi\sqrt{\frac{1-k}{1+k}}\ln\left(\frac{1-k}{1+k}\right)}{(\frac{1-k}{1+k}-1)(k^2-1)}\\
&=\frac{\pi\sqrt{\frac{1-k}{1+k}}\ln\left(\frac{1-k}{1+k}\right)}{(\frac{-2k}{1+k})(k^2-1)}\\
&=\frac{\pi\sqrt{\frac{1-k}{1+k}}\ln\left(\frac{1-k}{1+k}\right)}{2k(1-k)}\\
&=\boxed{\frac{\pi\ln\left(\frac{1-k}{1+k}\right)}{2k\sqrt{1-k^2}}}
\end{align}$
</s>