<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] Another method which gives a closed form result. Using the identity
\begin{equation}
\ln\frac{1-x}{1+x}=-2\operatorname{arctanh} x
\end{equation} 
and parity of the integrand, the integral can be written as
\begin{align}
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)\\
&=-\int_{-1}^1 \frac{\mathrm{d}x}{\sqrt{1-x^2}}\frac{x }{1-k^2x^2}\operatorname{arctanh} x\\
&=-\int_{-\infty}^\infty \frac{u\sinh u}{\cosh^2u-k^2\sinh^2u}du
\end{align}
The last integral is obtained by the substitution $x=\tanh u$. 
Then
\begin{align}
I(k)&=-\frac{1}{2k}\left[J(k)-J(-k)\right]\\
J(k)&=\int_{-\infty}^\infty \frac{udu}{\cosh u-k\sinh u}
\end{align}
To evaluate $J(k)$, we enforce the substitution $t=e^u$ and, denoting $m^2=\frac{1+k}{1-k}$, we obtain
\begin{align}
J(k)&=\frac{2}{1-k}\int_0^\infty \frac{\ln v}{v^2+m^2}dv\\
&=\frac{2}{m(1-k)}\int_0^\infty \frac{\ln mt}{t^2+1}dt\\
&=\frac{2}{m(1-k)}\left[\int_0^\infty \frac{\ln t}{t^2+1}dt+\int_0^\infty \frac{\ln m}{t^2+1}dt\right]
\end{align} 
We know that $\int_0^\infty \frac{\ln t}{t^2+1}dt=0$, thus
\begin{equation}
J(k)=\frac{\pi \ln m}{m(1-k)}=\frac{\pi}{2}\frac{\ln \frac{1+k}{1-k}}{\sqrt{1-k^2}}
\end{equation} 
Finally,
\begin{equation}
I(k)=\frac{\pi}{2k\sqrt{1-k^2}}\ln\frac{1- k}{1+ k}
\end{equation} 
</s>