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.

Another solution, this time using Mike Spivey's alternating Euler sum:
\begin{align*}
I &=-\int_0^1\ln(1-x)\ln(1+x)\,\mathrm dx \\ 
 &= \int_0^1\ln(1-x)\sum_{n=1}^\infty\frac{(-1)^n}{n}x^n\,\mathrm dx,\qquad\text{Mercator series}\\ 
 &= \sum_{n=1}^\infty\frac{(-1)^n}{n}\int_0^1\ln(1-x)x^n\,\mathrm dx\\ 
 &= \sum_{n=1}^\infty\frac{(-1)^n}{n}\cdot\frac{\partial}{\partial\alpha}\left[ \int_0^1(1-x)^\alpha x^n\,\mathrm dx\right ]\Bigg\vert_{\alpha=0}\\ 
 &= \sum_{n=1}^\infty\frac{(-1)^n}{n}\cdot\frac{\partial}{\partial\alpha} \text{B}\left(\alpha+1,\,n+1 \right )\Bigg\vert_{\alpha=0},\qquad\text{beta function}\\ 
 &= \sum_{n=1}^\infty\frac{(-1)^n}{n}\cdot\left[ \frac{\Gamma(\alpha+1)\Gamma(n+1)\left(\psi(\alpha+1)-\psi(\alpha+n+2) \right )}{\Gamma(\alpha+n+2)}\right ]\Bigg\vert_{\alpha=0}\\ 
 &= -\sum_{n=1}^\infty\frac{(-1)^n}{n}\cdot\frac{H_{n+1}}{n+1},\qquad\text{used }\psi(1)=-\gamma\text{ and }\psi(m)=H_{m-1}-\gamma\\ 
 &= -\sum_{n=1}^\infty\frac{(-1)^nH_{n+1}}{n}+\sum_{n=1}^\infty\frac{(-1)^nH_{n+1}}{n+1}\\ 
 &= -\sum_{n=1}^\infty\frac{(-1)^nH_n}{n}-\sum_{n=1}^\infty\frac{(-1)^n}{n(n+1)}+\sum_{n=1}^\infty\frac{(-1)^nH_{n+1}}{n+1}\\ 
 &= -\sum_{n=1}^\infty\frac{(-1)^nH_n}{n}-1+2\ln(2)+\sum_{n=1}^\infty\frac{(-1)^nH_{n+1}}{n+1}\\
 &= -2\sum_{n=1}^\infty\frac{(-1)^nH_n}{n}-2+2\ln(2)\\
 &= \frac{\pi^2}{6}+2\ln(2)-\ln^2(2)-2,\qquad\text{applied Mike's sum.}
\end{align*}
