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,
$$-\sum_{k=0}^{\infty}{1\over 2^{k+1}(1+k)^2}={\ln^2(2)-\zeta(2) \over 2}$$
 A: $$ \int_{0}^{1}\frac{\log x}{2-x}\,dx = \frac{1}{2}\sum_{n\geq 0}\int_{0}^{1}\left(\frac{x}{2}\right)^n\log(x)\,dx =-\sum_{n\geq 1}\frac{1}{2^n n^2}=-\text{Li}_2\left(\frac{1}{2}\right)$$
that can be computed through the dilogarithm reflection formula:
$$ \text{Li}_2(x)+\text{Li}_2(1-x) = \zeta(2)-\log(x)\log(1-x) $$
by substituting $x=\frac{1}{2}$.
A: start with $$\int_0^1 \ln(1-x) x^{n-1}dx = -\frac{H_{n}}{n}$$
change of variable and expand in power series
$$\int_0^1 \frac{\ln(x)}{2-x}dx = \int_0^1 \frac{\ln(1-x)}{1+x} = \sum_{n=0}^\infty (-1)^n \int_0^1\ln(1-x) x^n dx = \sum_{n=1}^\infty (-1)^n \frac{H_{n}}{n}$$
and use a trick found by  Jack D'Aurizio 
$$\frac{\ln^2(1-x)}{2} =  \int_0^x \frac{-\ln(1-t) }{1-t} dt = \sum_{k=1}^\infty\sum_{n=0}^\infty \int_0^x\frac{t^{n+k}}{k}dt = \sum_{k=1}^\infty\sum_{n=0}^\infty \frac{x^{n+k+1}}{k(n+k+1)}$$ $$ = \sum_{m= 2}^\infty \frac{x^{m}}{m} \sum_{k=1}^{m-1} \frac{1}{k} = \sum_{m =2}^\infty \frac{x^{m}}{m} H_{m-1}$$
hence
 $$\frac{\ln^2(2)}{2}  =\frac{\ln^2(1-(-1))}{2} = \sum_{m =2}^\infty \frac{(-1)^{m}}{m} H_{m-1} = 
 -\sum_{m=1}^\infty \frac{(-1)^m}{m^2} +\sum_{m=1}^\infty (-1)^{m}\frac{H_m}{m}  $$ $$= \eta(2) +\int_0^1 \frac{\ln(x)}{2-x}dx$$
i.e. with $\eta(2) = (1-2^{1-2})\zeta(2)$
$$\int_0^1 \frac{\ln(x)}{2-x}dx = \frac{\ln^2(2)}{2} - \frac{\zeta(2)}{2}$$

proof of the first formula : 
$\displaystyle \frac{1-x^n}{1-x} = \sum_{m=0}^{n-1} x^m\quad$  hence $\quad\displaystyle \int_0^1 \frac{1-x^n}{1-x} dx = \sum_{m=0}^{n-1} \frac{1}{m+1} = H_n\quad$ and integrating by parts $\quad\displaystyle \int_0^1 \frac{1-x^n}{1-x} dx = \lim_{a \to 1^-} (x^n-1)\ln(1-x)\mid_0^a $ $- \int_0^1 n x^{n-1} \ln(1-x)dx = -n\int_0^a  x^{n-1} \ln(1-x)dx$
A: 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,
$\boxed{I=\dfrac{1}{2}(\ln 2)^2-\dfrac{1}{2}\zeta(2)}$
