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.
 A: Naaah. Since $\frac{1}{1-x^2}=1+x^2+x^4+\ldots$ over $(0,1)$ and $\int_{0}^{1}x^{2n}\log(x)\,dx = -\frac{1}{(2n+1)^2}$,
$$ \int_{0}^{1}\frac{\log x}{1-x^2}\,dx = -\sum_{n\geq 0}\frac{1}{(2n+1)^2} = -\left[\zeta(2)-\tfrac{1}{4}\zeta(2)\right] $$
and $\lambda=\color{red}{8}$.
A: 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}
A: For another approach, take $f(x)=x^2$ on $[0,2\pi]$ and extend periodically. Then the Fourier series for $f$ converges to $\frac{1}{2}(\text{jump})$ where "jump" is the value of the difference of the left-and right-handed limit as $x\to 2\pi.$ For this $f$, jump=$4\pi^2$. Therefore, we have, after computing the Fourier series for $f$ and substituting $x=2\pi,$
$\frac{4\pi^{2}}{3}+\sum^{\infty}_{n=1}\frac{4\cos(2\pi n)}{n^2}=2\pi^2\Rightarrow\sum^{\infty}_{n=1}\frac{1}{n^2}=\frac{\pi^{2}}{6}$
A: 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
$$\begin{align}
\int_0^1 \frac{\log(x)}{1-x^2}\,dx&=\frac34 \int_0^1 \frac{\log(x)}{1-x}\,dx\\\\
&=\frac34 \int_0^1 \frac{\log(1-x)}{x}\,dx\\\\
&=\frac34\int_0^1 \frac{-\sum_{n=1}^\infty \frac{x^n}{n}}{x}\,dx\\\\
&=-\frac34\sum_{n=1}^\infty\frac1n \int_0^1 x^{n-1}\,dx\\\\
&=-\frac34\sum_{n=1}^\infty\frac1{n^2}\\\\
&-\frac{\pi^2}{8}
\end{align}$$
