Sum of reciprocals of odd squares and integral of $\log(x)/(1-x^2))$ I would like to prove that $\int_{0}^{1}\frac{\log(x))}{1-x^2}\,dx=-\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}. $
I know that the RHS is equal to $-\frac{\pi^2}{8} $.
Additionally, I have tried to make the substitution $x=\sin(u)$ to the integral, which leaves with $\int_{u=0}^{\pi/2} \frac{\log(\sin(x))}{\cos(x)}\,dx $ but I can't seem to proceed from here.
 A: Using partial fraction expansion reveals
$$\frac1{1-x^2}=\frac12\left(\frac{1}{1-x}+\frac1{1+x}\right)$$

Next using the Taylor series for $\log(x)$ and owing to uniform convergence, we can integrate term by term to arrive at
$$\begin{align}
\int_0^1 \frac{\log(x)}{1-x}\,dx&=-\sum_{n=1}^\infty (-1)^{n-1}\int_0^1 \frac{(x-1)^{n-1}}{n}\,dx\\\\
&=\sum_{n=1}^\infty  (-1)^{n-1}\frac{(-1)^{n}}{n^2}\tag1
\end{align}$$

Next, using integration by parts on the integral $\int_0^1\frac{\log(x)}{1+x}\,dx$ we find that
$$\begin{align}
\int_0^1 \frac{\log(x)}{1+x}\,dx&=-\int_0^1 \frac{\log(1+x)}{x}\,dx\\\\
&=-\sum_{n=1}^\infty (-1)^{n-1} \int_0^1 \frac{x^{n-1}}{n}\,dx\\\\
&=-\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^2}\tag2
\end{align}$$

Finally, adding $(1)$ and $(2)$ and dividing by $2$ yields
$$\begin{align}
\int_0^1 \frac{\log(x)}{1-x^2}\,dx&=-\frac12\sum_{n=1}^\infty \left(\frac{1+(-1)^{n-1}}{n^2}\right)\\\\
&=-\sum_{n=1}^\infty\frac1{(2n-1)^2}
\end{align}$$
as was to be shown!
A: Anothe solution
$$I=\int\frac{\log(x)}{1-x^2}\,dx=\sum_{n=0}^{\infty}\int x^{2n}\log(x)\,dx=\sum_{n=0}^{\infty}\frac{x^{2 n+1} ((2 n+1) \log (x)-1)}{(2 n+1)^2}$$
Using the bounds, the result.
