<s>[INST] 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. 
 [/INST] 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 $\text{Li}_{n}(\bar{z}) = \overline{\text{Li}_{n}(z)}$.
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