Help with $-\int_0^1 \ln(1+x)\ln(1-x)dx$ I have been attempting to evaluate this integral and by using wolfram alpha I know that the value is$$I=-\int_0^1 \ln(1+x)\ln(1-x)dx=\frac{\pi^2}{6}+2\ln(2)-\ln^2(2)-2$$
My Attempt:
I start off by parametizing the integral as $$I(a)=\int_0^1 -\ln(1+x)\ln(1-ax)dx$$
where $I=I(1)$. I then differentiate to get $$I'(a)=\int_0^1 \frac{ax\ln(1+x)}{1-ax}dx=\int_0^1 ax\ln(1+x)\sum_{n=0}^\infty(ax)^ndx=\sum_{n=1}^\infty a^{n+1}\int_0^1 x^{n+1}\ln(1+x)dx$$
Evaluating this integral by integration by parts and geometric series I get
$$\int_0^1 x^{n+1}\ln(1+x)dx=\frac{x^{n+2}}{n+2}\ln(1+x)|_0^1-\frac{1}{n+2}\int_0^1 \frac{x^{n+2}}{1+x}dx=\frac{\ln(2)}{n+2}-\frac{1}{n+2}\int_0^1 x^{n+2}\sum_{k=0}^\infty(-x)^kdx=\frac{\ln(2)}{n+2}-\frac{1}{n+2}\sum_{k=0}^\infty(-1)^k\int_0^1 x^{k+n+2}dx=\frac{\ln(2)}{n+2}-\frac{1}{n+2}\sum_{k=0}^\infty\frac{(-1)^k}{k+n+2}=\frac{\ln(2)}{n+2}-\frac{1}{2(n+2)}\left(\psi_0\left(\frac{n}{2}+2\right)-\psi_0\left(\frac{n}{2}+\frac{3}{2}\right)\right)$$ So I arrive at $$I'(a)=\sum_{n=0}^\infty a^{n+1}\left(\frac{\ln(2)}{n+2}-\frac{1}{2(n+2)}\left(\psi_0\left(\frac{n}{2}+2\right)-\psi_0\left(\frac{n}{2}+\frac{3}{2}\right)\right)\right)$$
Re-indexing I get
$$I'(a)=\frac{\ln(2)}{a}\sum_{n=2}^\infty \frac{a^n}{n}+\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n+1}{2}\right)}{n}a^{n-1}-\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n}{2}+1\right)}{n}a^{n-1}$$Integrating both sides from $0$ to $1$ I recover $I(1)$
$$I(1)=\int_0^1 \frac{\ln(2)}{a}\left(-\ln(1-a)-a\right)da+\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n+1}{2}\right)}{n^2}-\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n}{2}+1\right)}{n^2}$$ Then using the integral equation for the Dilogarithm I arrive at
$$I(1)=\ln(2)\int_0^1 -\frac{\ln(1-a)}{a}da-\ln(2)+\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n+1}{2}\right)}{n^2}-\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n}{2}+1\right)}{n^2}$$
$$I(1)=\frac{\ln(2)\pi^2}{6}-\ln(2)+\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n+1}{2}\right)}{n^2}-\frac{1}{2}\sum_{n=2}^\infty \frac{\psi_0\left(\frac{n}{2}+1\right)}{n^2}$$ 
At this point I could not continue further as I did not know how to simplify the Digamma terms in the sums. I think that by using the Digamma function's relation to the Harmonic Numbers it could be possible to exploit known values of Harmonic sums to arrive at the answer but I could not get the sums in a form where this would work. If anyone could help me continue further or let me know if I am on the right track I would greatly appreciate it. Thank you in advance.

Proceed as follows
\begin{align}
&\int_0^1 \ln(1-x)\ln(1+x)\,dx \\
=&\int_0^1\ln2 \ln(1-x)dx +\int_0^1 {\ln(1-x)\ln\frac{1+x}2dx}, \>\>\>{IBP} \\
=& -\ln2 +\int_0^1 x\left( \frac{\ln\frac{1+x}2}{1-x} -\frac{\ln(1-x)}{1+x}\right) dx\\=&-\ln2 +1 -\underbrace{\int_0^1 \ln\frac{1+x}2dx}_{=\ln2-1}+\int_0^1\frac{\ln(1-x)}{1+x}dx +\int_0^1  \underbrace{\frac{\ln\frac{1+x}2}{1-x}dx}_{IBP}\\
=& 2-2\ln2 +2 \int_0^1\frac{\ln(1-x)}{1+x}dx, \>\>\>\>\>
{t=\frac{1-x}{1+x}}\\
=&2-2\ln2 +2\int_0^1 \frac{\ln2-\ln(1+t)+\ln t}{1+t}dt\\
=& 2-2\ln2 +\ln^22-\frac{\pi^2}6
\end{align}
