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. 

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 $$I(a)=\int_0^1\frac{\ln(1+a^2y^2)}{y(1+y)}dy, \quad I(0)=0, \quad I(1)=\mathcal{J}$$
$$I'(a)=\int_0^1\frac{2ay}{(1+y)(1+a^2y^2)}dy=2\frac{\tan^{-1}(a)}{1+a^2}+\frac{a\ln(1+a^2)}{1+a^2}-\ln(2)\frac{2a}{1+a^2}$$
$$\Longrightarrow \mathcal{J}=2\int_0^1\frac{\tan^{-1}(a)}{1+a^2}da+\int_0^1\frac{a\ln(1+a^2)}{1+a^2}da-\ln(2)\int_0^1\frac{2a}{1+a^2}da\\=\left(\frac{\pi}{4}\right)^2+\frac14\ln^2(2)-\ln^2(2)=\boxed{\frac{\pi^2}{16}-\frac34\ln^2(2)}$$
