TITLE: If $ \int_{0}^{1} \frac{\ln x}{1-x^2} dx = -\frac{π^2}{\lambda} $ find $\lambda$ given that $\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{π^2}{6} $
QUESTION [6 upvotes]: If $$\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{π^2}{6} $$ then $$ \int_{0}^{1} \frac{\ln x}{1-x^2} dx = -\frac{π^2}{\lambda} $$ then the value of $\lambda$ equals?

My attempt- I tried using integrating by parts to the integral. As a result, I'm left with $$-\frac{1}{2} \int_{0}^{1} \frac{\ln\left(\frac{1+x}{1-x}\right)}{x} dx $$ Now I'm stuck in here! I don't know how to move on from here! Or maybe integrating by parts was a bad option? If it is, please guide me to a solution or Please help me on how to continue from here! 
Any help would be appreciated.

REPLY [2 votes]: First we use partial fraction expansion to write
$$\begin{align}
\int_0^1 \frac{\log(x)}{1-x^2}\,dx&=\frac12\int_0^1 \frac{\log(x)}{1-x}\,dx-\frac12\int_0^1 \frac{\log(x)}{1+x}\,dx\tag1
\end{align}$$
Next, enforcing the substitution $x\mapsto x^2$ reveals 
$$\begin{align}
\int_0^1 \frac{\log(x)}{1-x}\,dx&=\int_0^1 \frac{\log(x^2)}{1-x^2}\,2x\,dx\\\\
&=2\int_0^1 \frac{\log(x)}{1-x}\,dx+2\int_0^1 \frac{\log(x)}{1+x}\,dx \tag2
\end{align}$$
whence we find from $(2)$ that 
$$\int_0^1 \frac{\log(x)}{1+x}\,dx=-\frac12\int_0^1 \frac{\log(x)}{1-x}\,dx\tag3$$
Substituting $(3)$ into $(1)$ we obtain
$$\begin{align}
\int_0^1 \frac{\log(x)}{1-x^2}\,dx&=\frac34 \int_0^1 \frac{\log(x)}{1-x}\,dx\\\\
&=\frac34 \int_0^1 \frac{\log(1-x)}{x}\,dx\\\\
&=\frac34\int_0^1 \frac{-\sum_{n=1}^\infty \frac{x^n}{n}}{x}\,dx\\\\
&=-\frac34\sum_{n=1}^\infty\frac1n \int_0^1 x^{n-1}\,dx\\\\
&=-\frac34\sum_{n=1}^\infty\frac1{n^2}\\\\
&-\frac{\pi^2}{8}
\end{align}$$