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} $ 
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.

Use the expansion
$$\ln\frac{1+x}{1-x}=2\left(x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots\right)$$
then
\begin{align}
-\frac{1}{2} \int_{0}^{1} \frac{\ln\frac{1+x}{1-x}}{x} dx
&= -\frac{1}{2} \int_{0}^{1} \dfrac1x2\left(x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots\right) dx \\
&= -\left(1+\frac{1}{3^2}+\frac{1}{5^2}+\cdots\right) \\
&= \color{blue}{-\dfrac{\pi^2}{8}}
\end{align}
