$\int_{-1}^{1} \log\left(\frac{1+x}{1-x} \right) \frac{1}{1-ax} dx.$ 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.

The simple Mathematica result given by Varun Vejalla in the comments is indeed correct. In order to make the final result look a bit nicer, we can define the function $f \colon (-1,1) \to \mathbb{R}$,
$$ f(a) = \int \limits_{-1}^1 \frac{\operatorname{artanh}(x)}{1-ax} \, \mathrm{d} x \, . $$
Since $\operatorname{artanh}(x) = \frac{1}{2} \log\left(\frac{1+x}{1-x}\right)$, your integral is $2f(a)$.
We first let $x = \frac{1-t}{1+t} ~ \Leftrightarrow ~ t = \frac{1-x}{1+x}$ to obtain
$$ f(a) = \frac{1}{1+a} \int \limits_0^\infty \frac{-\log(t)}{(1+t) \left(\frac{1-a}{1+a} +t\right)} \, \mathrm{d} t \, .$$
Under $t \mapsto \frac{1-a}{1+a} t^{-1}$ this expression transforms into
$$ f(a) = \frac{1}{1+a} \int \limits_0^\infty \frac{\log(t) + 2 \operatorname{artanh}(a)}{(1+t) \left(\frac{1-a}{1+a} +t\right)} \, \mathrm{d} t \, .$$
Averaging these two equations we find (the final integral is easy to compute using partial fractions)
$$ f(a) = \frac{\operatorname{artanh}(a)}{1+a} \int \limits_0^\infty \frac{\mathrm{d} t}{(1+t) \left(\frac{1-a}{1+a} +t\right)} =  \frac{\operatorname{artanh}(a)}{1+a} \frac{\log\left(\frac{1-a}{1+a}\right)}{\frac{1-a}{1+a} - 1} = \frac{\operatorname{artanh}^2(a)}{a} $$
for $a \in (-1,1)$ (with $f(0) = 0$).
In particular, $2 f\!\left(\frac{1}{2}\right) = \log^2(3)$, which agrees with Eevee Trainer's result after some dilogarithm identities are used.
