Prove $\int_0^1\frac{\ln2-\ln\left(1+x^2\right)}{1-x}\,dx=\frac{5\pi^2}{48}-\frac{\ln^22}{4}$ How does one prove the following integral

\begin{equation}
\int_0^1\frac{\ln2-\ln\left(1+x^2\right)}{1-x}\,dx=\frac{5\pi^2}{48}-\frac{\ln^22}{4} 
\end{equation}

Wolfram Alpha and Mathematica can easily evaluate this integral. This integral came up in the process of finding the solution this question: Evaluating $\displaystyle\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}}{n}$. There are some good answers there but avoiding this approach. I have been waiting for a day hoping an answers would be posted using this approach, but nothing shows up. The integral cannot be evaluated separately since each terms doesn't converge. I tried integration by parts but the problem arises when substituting the bounds of integration.
I would appreciate if anyone here could provide an answer where its approach using integral only preferably with elementary ways. 

In case the OP is curious about the harmonic series, here is an easy approach:
Using the classical identity 
$$\sum_{n=1}^\infty (-1)^{n-1} f(2n)=-\Re\sum_{n=1}^\infty i^n f(n)$$
then 
$$S=\sum_{n-1}^\infty (-1)^{n-1}\frac{H_{2n}}{n}=2\sum_{n-1}^\infty (-1)^{n-1}\frac{H_{2n}}{2n}=-2\Re\sum_{n=1}^\infty i^n\frac{H_n}{n}$$
Using the generating function 
$$\sum_{n=1}^\infty x^n\frac{H_n}{n}=\frac12\ln^2(1-x)+\operatorname{Li}_2(x)$$
Set $x=i$ we get
$$S=-2\Re\left(\frac12\ln^2(1-i)+\operatorname{Li}_2(i)\right)$$
$$=-2\Re\left(-\frac{5\pi^2}{96}-\frac18\ln^2(2)\right)=\frac{5\pi^2}{48}-\frac14\ln^2(2)$$
