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. 

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\begin{align}
&\bbox[10px,#ffd]{\int_{0}^{1}{\ln\pars{2} - \ln\pars{1 + x^{2}} \over
1 - x}\,\dd x}
\\[5mm] = &\
-\int_{x\ =\ 0}^{x\ =\ 1}\bracks{\ln\pars{2} -
\ln\pars{1 + x^{2}}}\,\dd\ln\pars{1 - x}
\\[5mm] \stackrel{\mrm{IBP}}{=}\,\,\, &
\int_{0}^{1}\ln\pars{1 - x}\pars{-\,{2x \over 1 + x^{2}}}\,\dd x =
-2\int_{0}^{1}{x\ln\pars{1 - x} \over \pars{x + \ic}\pars{x - \ic}}\,\dd x
\\[5mm] = &\
-\int_{0}^{1}\ln\pars{1 - x}\pars{{1 \over x + \ic} +
{1 \over x - \ic}}\,\dd x =
-\,\Re\int_{0}^{1}{\ln\pars{1 - x} \over x + \ic}\,\dd x
\\[5mm] = &\
-\,\Re\int_{0}^{1}{\ln\pars{x} \over 1 + \ic - x}\,\dd x =
-\,\Re\int_{0}^{1}{\ln\pars{x} \over 1 - x/\pars{1 + \ic}}
\,{\dd x \over 1 + \ic}
\\[5mm] = &\
-\,\Re\int_{0}^{\pars{1\ -\ \ic}/2}{\ln\pars{\bracks{1 + \ic}x} \over
1 - x}\,\dd x
\\[5mm] \stackrel{\mrm{IBP}}{=}\,\,\,&
-2\,\Re\int_{0}^{\pars{1\ -\ \ic}/2}{\ln\pars{1 - x} \over x}\,\dd x
\\[5mm] = &\
2\,\Re\int_{0}^{\pars{1\ -\ \ic}/2}\mrm{Li}_{2}'\pars{x}\,\dd x =
2\,\Re\,\mrm{Li}_{2}\pars{1 - \ic \over 2}
\\[5mm] = &\
\mrm{Li}_{2}\pars{{1 \over 2} - {1 \over 2}\,\ic} +
\mrm{Li}_{2}\pars{{1 \over 2} + {1 \over 2}\,\ic}
\\[5mm] = &
{\pi^{2} \over 6} - \ln\pars{{1 \over 2} - {1 \over 2}\,\ic}
\ln\pars{{1 \over 2} + {1 \over 2}\,\ic}
\label{1}\tag{1}
\end{align}
In (\ref{1}) I used the
Euler Reflection Formula.

Then,
\begin{align}
&\bbox[10px,#ffd]{\int_{0}^{1}{\ln\pars{2} - \ln\pars{1 + x^{2}} \over
1 - x}\,\dd x} =
{\pi^{2} \over 6} - 
\verts{-\,{1 \over 2}\,\ln\pars{2} - {1 \over 4}\,\pi\ic}^{\, 2}
\\[5mm] = &\
\bbx{{5\pi^{2} \over 48} - {1 \over 4}\,\ln^{2}\pars{2}}
\approx 0.9080
\end{align}
