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

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.

Just a preface but I feel this is an incomplete answer, in that my derivativion ultimately just gets to a series and one I feel there is no clear simplification for and hence no closed answer (via this method). However, I feel there probably is a closed-form answer if you don't mind special functions, based on how $a=1/2$ goes.

So treat this more as an extended comment, I suppose.

First, note that $x \in [-1,1], a \in (0,1) \implies |ax| < 1$ so we may expand into a geometric series,

$$\mathcal{I} := \int_{-1}^1 \log \left( \frac{1+x}{1-x} \right) \frac{1}{1-ax} \, \mathrm{d} x = \int_{-1}^1 \log \left( \frac{1+x}{1-x} \right) \sum_{k \ge 0} a^k x^k \, \mathrm{d} x $$

On the assumption we can do so, let's bring the summation and $a^k$ outside:

$$ \mathcal{I} = \sum_{k \ge 0} a^k \int_{-1}^1 \log \left( \frac{1+x}{1-x} \right) x^k \, \mathrm{d} x$$

It should not be difficult to see that our integrand is even for $k$ odd, and odd for $k$ even, allowing us to claim

$$ \mathcal{I} = \sum_{k \ge 0} 2a^{2k+1} \int_{0}^1 \log \left( \frac{1+x}{1-x} \right) x^{2k+1} \, \mathrm{d} x$$

Expanding the logarithm into two by its quotient properties, and handling each as a series and combining them ($x$ is within the permissible range), we get

$$ \log \left( \frac{1+x}{1-x} \right) = 2 \sum_{m \ge 0} \frac{x^{2m+1}}{2m+1}$$

and so

$$\mathcal{I} = \sum_{k \ge 0} 2a^{2k+1} \int_0^1 x^{2k+1} \cdot 2 \sum_{m \ge 0} \frac{x^{2m+1}}{2m+1} \, \mathrm{d} x$$

From here, again, we make the assumption we can bring out the summation in $m$, the constants, and then get

$$\mathcal{I} = \sum_{k \ge 0} \sum_{m \ge 0} \frac{4}{2m+1} a^{2k+1} \int_0^1 x^{2(k+m+1)} \, \mathrm{d} x = \sum_{k \ge 0} \sum_{m \ge 0} \frac{4}{(2m+1)(2k+2m+3)} a^{2k+1} $$

Comments:

• Some attempts at approximations using Wolfram seem to match pretty well. Using upper limits of $300$ for $m,k$, we get the following errors:

  • $a=1/2$ gives an error of $ -0.00220995$ (link)
  • $a=1/e$ gives an error of $ -0.00141078$ (link)
  • $a=1/3$ gives an error of $ -0.00124352$ (link)
  • $a=1/4$ gives an error of $-0.000884369$ (link)
  • $a=1/10$ gives an error of $-0.00033502$ (link)

• Curiously, Wolfram can give an exact answer (if you don't mind polylogarithms) for the case of $a=1/2$, giving $$\int_{-1}^1 \log \left( \frac{1+x}{1-x} \right) \frac{1}{1-x/2} = 2 \operatorname{Li}_2 \left( - \frac 1 2 \right) - 2 \operatorname{Li}_2 \left( \frac 2 3 \right) + \frac{π^2}{3}+ \log^2(2)≈1.20695$$ (Link here.) Hence why I think this answer is incomplete and maybe there is a closed form for all $a \in (0,1)$. However, hopefully it's helpful; who knows, maybe some kind of series manipulation can be done.

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, $$\int_{-1}^{1} \frac{\log \left(\frac{1+x}{1-x} \right) }{1-ax} \, \mathrm dx = \frac{\log^{2} \left(\frac{1+a}{1-a} \right)}{2a} =\frac{2 \operatorname{artanh}^{2}(a)}{a}. $$