$\int_{-1}^{1} \log\left(\frac{1+x}{1-x} \right) \frac{1}{1-ax} dx.$ 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: 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.
A: 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: 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, $$\int_{-1}^{1} \frac{\log \left(\frac{1+x}{1-x} \right) }{1-ax} \, \mathrm dx = \frac{\log^{2} \left(\frac{1+a}{1-a} \right)}{2a} =\frac{2 \operatorname{artanh}^{2}(a)}{a}.  $$
A: Another approach is to use the following integral representation of the natural logarithm:
$$\log\left(\lambda\right) = \int_{0}^{\infty} \left( \frac{1}{1+t} - \frac{1}{\lambda+t} \right) \, dt$$
$$\implies I=\int_{-1}^{1} \int_{0}^{\infty} \frac{2x}{(1+t)((x-1)t-x-1)(a x-1)} \, dt \, dx$$
Interchanging the order of integration, and using partial fractions to integrate the integrand with respect to x and subtracting the limits yields:
$$I=\int_{0}^{\infty} \frac{4(t-1) \, \text{arctanh} \, (a) -2a(t+1) \log (t)}{a(t(a-1)+a+1)(t^2-1)} \, dt$$
This can be integrated fairly easily to give a polylogarithm and some logarithm terms, and after computing the limits gives:
$$I = -\frac{\log^2(a-1)-\log^2(a+1)+2\log(1-a)(\log(a+1)-\log(a-1))+\pi^2}{2a}$$
This admits a rather interesting identity that can be confirmed with polylogarithm identities:
$$\log^2\left(\frac{2}{1+a} - 1\right) = -\log^2(a-1)+\log^2(a+1)-2\log(1-a)(\log(a+1)-\log(a-1))-\pi^2$$
A: Note that
$$\frac{d}{dx}\text{Li}_2\left(\frac{1-ax}{1+a}\right)=\frac{a\ln\left(\frac{a}{1+a}\right)+a\ln(1+x)}{1-ax},$$
$$\frac{d}{dx}\text{Li}_2\left(\frac{1-ax}{1-a}\right)=\frac{a\ln\left(\frac{a}{a-1}\right)+a\ln(1-x)}{1-ax}.$$
On subtracting, we have
$$\frac{d}{dx}\left(\text{Li}_2\left(\frac{1-ax}{1+a}\right)-\text{Li}_2\left(\frac{1-ax}{1-a}\right)\right)=\frac{a}{1-ax}\left(\ln\left(\frac{a-1}{1+a}\right)+\ln\left(\frac{1+x}{1-x}\right)\right).$$
Integrating both sides,
$$\text{Li}_2\left(\frac{1-ax}{1+a}\right)-\text{Li}_2\left(\frac{1-ax}{1-a}\right)=-\ln\left(\frac{a-1}{1+a}\right)\ln(1-ax)+a\int\frac{\ln\left(\frac{1+x}{1-x}\right)}{1-ax}dx.$$
Then
$$\int_{-1}^1\frac{\ln\left(\frac{1+x}{1-x}\right)}{1-ax}dx=\left.\frac{\text{Li}_2\left(\frac{1-ax}{1+a}\right)-\text{Li}_2\left(\frac{1-ax}{1-a}\right)+\ln\left(\frac{a-1}{1+a}\right)\ln(1-ax)}{a}\right]_{-1}^1$$
$$=\frac{\text{Li}_2\left(\frac{1-a}{1+a}\right)+\text{Li}_2\left(\frac{1+a}{1-a}\right)+\ln\left(\frac{a-1}{1+a}\right)\ln\left(\frac{1-a}{1+a}\right)-2\zeta(2)}{a}.$$
By setting $z=\frac{a-1}{1+a}$ in the dilogarithm inversion formula:
$$\text{Li}_2(-z)+\text{Li}_2(-1/z)=-\frac12 \ln^2(z)-\zeta(2),$$
we get
$$\text{Li}_2\left(\frac{1-a}{1+a}\right)+\text{Li}_2\left(\frac{1+a}{1-a}\right)=-\frac12\ln^2\left(\frac{a-1}{1+a}\right)-\zeta(2).$$
Thus,
$$\int_{-1}^1\frac{\ln\left(\frac{1+x}{1-x}\right)}{1-ax}dx=\frac{-\frac12\ln^2\left(\frac{a-1}{1+a}\right)+\ln\left(\frac{a-1}{1+a}\right)\ln\left(\frac{1-a}{1+a}\right)-3\zeta(2)}{a}=\frac{\ln^2\left(\frac{1-a}{1+a}\right)}{2a}.$$
The last equality follows from writing $\ln\left(\frac{a-1}{1+a}\right)=i\pi+\ln\left(\frac{1-a}{1+a}\right)$, which follows from $\ln(-z)=i\pi+\ln(z)$.
