Want help solving $\int_0^1\frac{\ln(1+x)}{\ln(1-x)}dx$ When attempting to solve
$$\int_0^1\frac{\ln(1+x)}{\ln(1-x)}dx$$
I attempted using the Taylor series expansion of $\ln(1+x)$ first, but that came to no avail. I also tried splitting it up a bit with integration by parts by multiplying top and bottom by x, and then integrating $\frac{\ln(1+x)}{x}$ and differentiating $\frac{x}{\ln(1-x)}$, but I had no clue what to do with the functions involving $\mathrm{Li_2}(-x)$. I know the integral converges I'm just not sure how else to go about solving it.
 A: You can get an exponentially convergent series in the following:
\begin{align}
\int_0^1 \frac{\log(1+x)}{\log(1-x)} \, {\rm d}x &= \int_0^1 \frac{\log(2-x)}{\log(x)} \, {\rm d}x \\
&=\log(2-x) {\rm li}(x) \Big|_0^1 + \int_0^1 \frac{{\rm li}(x)}{2-x} \, {\rm d}x \\
&=\frac{1}{2} \sum_{n=0}^\infty \frac{1}{2^n}\int_0^1 {\rm li}(x) x^n \, {\rm d}x \\
&=-\sum_{n=0}^\infty \frac{\log(2+n)}{(n+1)\,2^{n+1}} \, .
\end{align}
A: You can have reasonable approximations expanding the integrand as a series built around $x=\frac 12$.
$$\left(
\begin{array}{cc}
 n & \text{result} \\
 1 & -0.584963 \\
 3 & -0.592170 \\
 5 & -0.590350 \\
 7 & -0.589367 \\
 9 & -0.588835 \\
 11 & -0.588514 \\
 13 & -0.588302 \\
 15 & -0.588153 \\
\cdots & \cdots \\
\infty & -0.587413
\end{array}
\right)$$
Another possible approach would be to write
$$\frac 1 {\log(1-x)}=\sum_{n=-1}^p a_n\, x^n$$ and to face integrals
$$I_n=\int_0^1 x^n \log(1+x)\,dx=\frac{(n+1) \left(H_{\frac{n}{2}}-H_{\frac{n-1}{2}}\right)+2(n+1)\log(2)-2 } {2(n+1)^2}$$
A: If we take the series for $\ln(1 + x)$, i.e. the famous "Mercator series"
$$\ln(1 + x) = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^n}{n}$$
then we have
$$\frac{\ln(1 + x)}{\ln(1 - x)} = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \frac{x^n}{\ln(1 - x)}$$
Now it remains to find the integrals
$$I_n := \int_{0}^{1} \frac{x^n}{\ln(1 - x)}\ dx$$
Here I don't have a really good solution right now (trying to antidifferentiate them can be done by $u = 1 - x$ and then the binomial theorem to get a sum of exponential integrals Ei, and I'm not sure how to simplify all that or if it can be done), but some experimentation with Wolfram suggests that the terms can be given with elementary expressions
$$\begin{align}
I_1 &= -\ln 2 = \ln\left(\frac{1}{2}\right)\\
I_2 &= -\ln\left(\frac{4}{3}\right) = \ln\left(\frac{3}{4}\right)\\
I_3 &= -\ln\left(\frac{32}{27}\right) = \ln\left(\frac{27}{32}\right)\\
I_4 &= -\ln\left(\frac{4096}{3645}\right) = \ln\left(\frac{3645}{4096}\right)\\
&\cdots
\end{align}$$
Hence we can at least get the following weird series formula:
$$\int_{0}^{1} \frac{\ln(1 + x)}{\ln(1 - x)}\ dx  = \ln\left(\frac{1}{2}\right) - \frac{1}{2} \ln\left(\frac{3}{4}\right) + \frac{1}{3} \ln\left(\frac{27}{32}\right) - \frac{1}{4} \ln\left(\frac{3645}{4096}\right) + \cdots$$
with the peculiar sequence of rational numbers
$$\frac{1}{2}, \frac{3}{4}, \frac{27}{32}, \frac{3645}{4096}, \cdots$$
within them.
ADD: Apparently, OEIS.org has some:
https://oeis.org/search?q=1%2C3%2C27%2C+3645&sort=&language=english&go=Search
which suggest that 
$$\int_{0}^{1} \frac{\ln(1 + x)}{\ln(1 - x)}\ dx = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \ln\left(\frac{1}{\prod_{k=1}^{n+1} k^{(-1)^{k} \binom{n}{k-1}}}\right)$$
ADD 2: We can then, of course, use log identities to get
$$\int_{0}^{1} \frac{\ln(1 + x)}{\ln(1 - x)}\ dx = \sum_{n=1}^{\infty} \sum_{k=1}^{n+1} \frac{(-1)^{n+k}}{n} \binom{n}{k-1} \ln(k)$$.
but not sure if there's anything simpler than that, or any finitary expressions.
