Ideas on approaching the integral $\int_0^1 \frac{\operatorname{artanh} x}{x} \mathrm d x$ I stumbled across this integral while trying to prove, $\displaystyle \sum_{n=0}^\infty \frac 1 {(2n + 1)^2} = \frac{\pi^2} 8 $, a brief sketch of my method being, 
$\displaystyle \sum_{n=0}^\infty \frac 1 {(2n + 1)^2} = \sum_{n=0}^\infty \int_0^1 x^{2n} \mathrm dx \int_0^1 y^{2n} \mathrm dy = \int_0^1 \int_0^1\sum_{n=0}^\infty (x^2 y^2)^n \mathrm d x \mathrm d y = \int_0^1 \int_0^1 \frac 1 {1 - x^2y^2}\mathrm d x \mathrm d y = \frac 1 2 \int_0^1 \frac 1 {x} \log\left(\frac{1+x}{1-x}\right) \mathrm dx = \int_0^1 \frac{\operatorname{artanh}(x)} x \mathrm d x$
Obviously, there's no elementary antiderivative, and I don't think using Wolfram to calculate dilogs would constitute a nice proof. I'm absolutely lost, so I would appreciate a nudge in the right direction. (maybe residues?)
Edit: looking for a method that doesn't assume or compute the result $\zeta(2) = \frac{\pi^2}{6}$

By integration by parts, one has
\begin{eqnarray}
\int_0^1 \frac{\operatorname{artanh} x}{x} \mathrm d x&=&\int_0^1\operatorname{artanh} x \mathrm d \ln x\\
&=&\operatorname{artanh} x\ln x|_0^1-\int_0^1\ln x\mathrm d\operatorname{artanh} x\\
&=&-\int_0^1\frac{\ln x}{1-x^2}dx.
\end{eqnarray}
The calculation of this integral follows this post.
