<s>[INST] $\int_{-1}^{1} \log\left(\frac{1+x}{1-x} \right) \frac{1}{1-ax} dx.$ 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.
 [/INST] 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.
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