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. 
 A: We have:
$$\begin{eqnarray*} I &=& -\int_{0}^{1}\frac{2x \log(1-x)}{1+x^2}=\sum_{k=1}^{+\infty}\frac{1}{k}\int_{0}^{1}\frac{2x^{k+1}}{1+x^2}\,dx=\sum_{k=1}^{+\infty}\frac{2}{k}\sum_{j=1}^{+\infty}\frac{(-1)^{j+1}}{k+2j}\\&=&\sum_{j=1}^{+\infty}\frac{(-1)^{j+1}H_{2j}}{j}.\end{eqnarray*}\tag{1}$$
Since
$$\sum_{j=1}^{+\infty}\left(\frac{1}{2j-1}+\frac{1}{2j}\right)x^{j-1} = \frac{\operatorname{arctanh}\sqrt{x}}{\sqrt{x}}-\frac{\log(1-x)}{2x}$$
we have:
$$\sum_{j=1}^{+\infty} H_{2j}\, x^{j-1} = \frac{\operatorname{arctanh}\sqrt{x}}{(1-x)\sqrt{x}}-\frac{\log(1-x)}{2x(1-x)},$$
$$\sum_{j=1}^{+\infty} (-1)^{j+1}H_{2j}\, x^{j-1} = \frac{\arctan\sqrt{x}}{(1+x)\sqrt{x}}+\frac{\log(1+x)}{2x(1+x)},\tag{2}$$
hence:
$$ I = \int_{0}^{1}\frac{\arctan\sqrt{x}}{(1+x)\sqrt{x}}+\frac{\log(1+x)}{2x(1+x)}\,dx=\int_{0}^{1}\frac{2\arctan x}{1+x^2}\,dx+\int_{0}^{1}\frac{\log(1+x)}{2x(1+x)}\,dx,$$
$$ I = \frac{\pi^2}{16}+\int_{0}^{1}\frac{\log(1+x)}{2x}\,dx-\int_{0}^{1}\frac{\log(1+x)}{2(1+x)}\,dx\tag{3}$$
and finally:

$$ I = \frac{\pi^2}{16}+\frac{\pi^2}{24}-\frac{\log^2 2}{4}.$$

A: As M.N.C.E. stated in the comments, integrating by parts shows that $$ \begin{align} \int_{0}^{1} \frac{\log(2) - \log(1+x^{2}) }{1-x} \ dx&= - 2 \int_{0}^{1} \frac{x \log(1-x)}{x^{2}+1} \ dx \\ &= - 2 \ \text{Re} \int_{0}^{1} \frac{\log(1-x)}{x+i} \ dx . \end{align}$$
Letting $ \displaystyle t = \frac{x+i}{1+i} $,
$$ \begin{align} \int  \frac{\log(1-x)}{x+i} \ dx &= \int \frac{\log [(1+i)(1-t)]}{t} \ dt \\ &= \log(1+i) \int \frac{dt}{t} + \int\frac{\log(1-t)}{t} \ dt \ \ \left(- \pi < \frac{\pi}{4} + \text{Arg} (1-t) \le \pi \right) \\ &= \log(1+i) \log(t) - \text{Li}_{2} (t) + C  \\ &= \log(1+i) \log \left(\frac{x+i}{1+i} \right) - \text{Li}_{2} \left(\frac{x+i}{1+i} \right) + C . \end{align}$$
Therefore,
$$ \begin{align}  &\int_{0}^{1} \frac{\log(2) - \log(1+x^{2}) }{1-x}  \\ &= - 2 \ \text{Re} \left[- \text{Li}_{2} (1) - \log(1+i) \log \left(\frac{1+i}{2}  \right) + \text{Li}_{2} \left(\frac{1+i}{2} \right) \right] \\ &= - 2 \ \text{Re} \left[- \frac{\pi^{2}}{6} - \left(\frac{\log 2}{2} + \frac{i \pi}{4} \right) \left(- \frac{\log 2}{2} + \frac{i \pi}{4} \right) \right] -2 \ \text{Re} \ \text{Li}_{2} \left(\frac{1+i}{2} \right)  \\ & = - 2 \left(\frac{\log^{2}(2)}{4} - \frac{5 \pi^{2}}{48} \right) - 2 \left(\frac{5 \pi^{2}}{96} - \frac{\log^{2} (2)}{8} \right) \\  &= \frac{5 \pi^{2}}{48} - \frac{\log^{2}(2)}{4} . \end{align}$$
To show that $$ \text{Re}  \  \text{Li}_{2} \left(\frac{1+i}{2} \right) = \frac{5 \pi^{2}}{96} - \frac{\log^{2} (2)}{8}$$ combine the reflection formula for the dilogarithm (5) with the property $\text{Li}_{n}(\bar{z}) = \overline{\text{Li}_{n}(z)}$.
A: Start with subbing $x=\frac{1-y}{1+y}$
$$\mathcal{I}=\int_0^1\frac{\ln(2)-\ln(1+x^2)}{1-x}dx=\int_0^1\frac{2\ln(1+y)-\ln(1+y^2)}{y(1+y)}dy$$
$$=2\int_0^1\frac{\ln(1+y)}{y}dy-2\int_0^1\frac{\ln(1+y)}{1+y}dy-\int_0^1\frac{\ln(1+y^2)}{y(1+y)}dy$$
$$=-2\operatorname{Li}_2(-y)|_0^1-\ln^2(1+y)|_0^1-\mathcal{J}$$
$$=\frac{\pi^2}{6}-\ln^2(2)-\left(\frac{\pi^2}{16}-\frac34\ln^2(2)\right)=\boxed{\frac{5\pi^2}{48}-\frac14\ln^2(2)}$$
Where the last integral $\mathcal{J}=\int_0^1\frac{\ln(1+y^2)}{y(1+y)}dy$ can be proved using the common Feynman's method:
Let $$I(a)=\int_0^1\frac{\ln(1+a^2y^2)}{y(1+y)}dy, \quad I(0)=0, \quad I(1)=\mathcal{J}$$
$$I'(a)=\int_0^1\frac{2ay}{(1+y)(1+a^2y^2)}dy=2\frac{\tan^{-1}(a)}{1+a^2}+\frac{a\ln(1+a^2)}{1+a^2}-\ln(2)\frac{2a}{1+a^2}$$
$$\Longrightarrow \mathcal{J}=2\int_0^1\frac{\tan^{-1}(a)}{1+a^2}da+\int_0^1\frac{a\ln(1+a^2)}{1+a^2}da-\ln(2)\int_0^1\frac{2a}{1+a^2}da\\=\left(\frac{\pi}{4}\right)^2+\frac14\ln^2(2)-\ln^2(2)=\boxed{\frac{\pi^2}{16}-\frac34\ln^2(2)}$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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}
A: 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
$$S=-2\Re\left(\frac12\ln^2(1-i)+\operatorname{Li}_2(i)\right)$$
$$=-2\Re\left(-\frac{5\pi^2}{96}-\frac18\ln^2(2)\right)=\frac{5\pi^2}{48}-\frac14\ln^2(2)$$
