Prove that $\int_{0}^{1}{\ln(x) \over 2-x}\mathrm dx={\ln^2(2)-\zeta(2)\over 2}$ $$I=\int_{0}^{1}{\ln(x) \over 2-x}\,\mathrm{d}x={\ln^2(2)-\zeta(2)\over 2}$$
$$(2-x)^{-1}={1\over2}\left(1-{x\over 2}\right)^{-1}$$
Using binomial series here
$$(2-x)^{-1}={1\over2}\left(1-{x\over 2}\right)^{-1}={1\over2}+{x\over 4}+{x^2\over8}+{x^3\over 16}+\cdots=\sum_{k=0}^{\infty}{x^k\over2^{k+1}}$$
$$I=\sum_{k=0}^{\infty}{1\over 2^{k+1}}\int_{0}^{1}x^k\ln(x)\,\mathrm{d}x$$
$$\int_{0}^{1}x^k\ln(x)\,\mathrm{d}x=-{1\over (1+k)^2}$$
$$I=-\sum_{k=0}^{\infty}{1\over 2^{k+1}(1+k)^2}$$
So this is where I got to, unable to determine this infinite sum. Do anyone know how to prove that,
$$-\sum_{k=0}^{\infty}{1\over 2^{k+1}(1+k)^2}={\ln^2(2)-\zeta(2) \over 2}$$

On the path of  user1952009,
Perform the change of variable $y=1-x$,
$\displaystyle I=\int_0^1 \dfrac{\ln(1-x)}{1+x}dx$
Perform the change of variable $y=\dfrac{1-x}{1+x}$,
$\begin{align}
I&=\int_0^1 \dfrac{\ln\left(\tfrac{2x}{1+x}\right)}{1+x}dx\\
&=\int_0^1 \dfrac{\ln 2}{1+x}dx+\int_0^1 \dfrac{\ln x}{1+x}dx-\int_0^1 \dfrac{\ln(1+x)}{1+x}dx\\
&=\ln 2\Big[\ln(1+x)\Big]_0^1+\int_0^1 \dfrac{\ln x}{1+x}dx-\Big[\dfrac{1}{2}(\ln(1+x))^2\Big]_0^1\\
&=\dfrac{1}{2}(\ln 2)^2+\int_0^1 \dfrac{\ln x}{1+x}dx
\end{align}$
$\begin{align}\int_0^1 \dfrac{\ln x}{1+x}dx&=\int_0^1 \ln x\left(\sum_{n=0}^{\infty} (-1)^n x^n\right)dx\\
&=\sum_{n=0}^{\infty} \left(\int_0^1  (-1)^n x^n\ln x dx\right)\\
&=-\sum_{n=0}^{\infty} \dfrac{(-1)^n}{(1+n)^2}\\
&=\sum_{n=1}^{\infty} \dfrac{(-1)^n}{n^2}\\
&=\sum_{n=1}^{\infty} \dfrac{1}{(2n)^2}-\sum_{n=0}^{\infty} \dfrac{1}{(2n+1)^2}\\
&=\dfrac{1}{4}\zeta(2)-\left(\zeta(2)-\dfrac{1}{4}\zeta(2)\right)\\
&=-\dfrac{1}{2}\zeta(2)
\end{align}$
therefore,
$\boxed{I=\dfrac{1}{2}(\ln 2)^2-\dfrac{1}{2}\zeta(2)}$
