$\int_{-1}^{1} \log\left(\frac{1+x}{1-x} \right) \frac{1}{1-ax} dx.$ Definite Integral: $$\int_{-1}^{1} \log\left(\frac{1+x}{1-x} \right) \frac{1}{1-ax} dx,$$
where $0 < a <1$.

I tried with integration by parts taking $\log\left(\frac{1+x}{1-x} \right) $ as the first function and $\frac{1}{1-ax} $ as the second function, but it did not work.
Need some help to compute it.

A complex analysis approach is to integrate the function $$f(z) = \frac{\left(\log(z+1) - \log(z-1)\right)^{\color{red}{2}}}{1-az}, \quad 0< a< 1, $$ where $0 \le \arg(z+1), \arg(z-1) < 2 \pi$,  around a dog bone contour that goes around branch cut on $[-1, 1]$.
(The nicest picture I can find of the contour in this answer.)
Since $f(z) \sim \mathcal{O}\left(\frac{1}{z^{2}} \right)$ as $|z| \to \infty$,  the residue of $f(z)$ at $\infty$ is zero.
Integrating clockwise around the contour, we get $$ \begin{align} \oint f(z) \, \mathrm dz &=\small \int_{-1}^{1} \frac{\left(\log(1+x)-\log(1-x) - \pi i \right)^{2}}{1-ax} \, \mathrm dx + \int_{1}^{-1}\frac{\left(\log(1+x)+ 2 \pi i - \log(1-x) - \pi i\right)^{2}}{1-ax} \mathrm dx\\ &=  -4 \pi i \int_{-1}^{1} \frac{\log \left(\frac{1+x}{1-x} \right) }{1-ax} \, \mathrm dx  \\ &= 2 \pi i \operatorname{Res}\left[f(z), \frac{1}{a}\right] \\ & = -2 \pi i \, \frac{\log^{2} \left(\frac{\frac{1}{a}+1}{\frac{1}{a}-1} \right)}{a}  \\&=-2 \pi i \, \frac{\log^{2} \left(\frac{1+a}{1-a} \right)}{a} . \end{align}$$
Therefore, $$\int_{-1}^{1} \frac{\log \left(\frac{1+x}{1-x} \right) }{1-ax} \, \mathrm dx = \frac{\log^{2} \left(\frac{1+a}{1-a} \right)}{2a} =\frac{2 \operatorname{artanh}^{2}(a)}{a}.  $$
