How find this integral $I=\int_{0}^{1}\frac{1}{2-x}\ln{\frac{1}{x}}dx$ Find this integral
$$I=\int_{0}^{1}{1 \over 2 - x}\,\ln\left(1 \over x\right)\,{\rm d}x$$
My idea: let $1-x=t$, then
$$I=\int_{0}^{1}{\ln\left(1 - t\right) \over 1 + t}\,{\rm d}t$$
 A: We have
$$I=\int_0^1\sum_{n=0}^\infty\frac{x^n}{2^{n+1}}\ln (1/x)\,dx
=\sum_{n=0}^\infty\frac{1}{2^{n+1}}\int_0^1x^n\ln(1/x)dx
=\sum_{n=0}^\infty\frac{1}{2^{n+1}(n+1)^2}$$
That is $$I=\sum_{n=1}^\infty\frac{1}{2^nn^2}=\hbox{Li}_2\left(\frac{1}{2}\right)=
\frac{\pi^2}{12} - \frac{\ln^22}{2}.$$
For more information on the Dilogarithm $\hbox{Li}_2$ see here.
A: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$
$\ds{I \equiv \int_{0}^{1}{1 \over 2 - x}\ln\pars{1 \over x}\,\dd x:\ {\large ?}}$

\begin{align}
\color{#00f}{\large I}&=-\,\half\int_{0}^{1}{\ln\pars{x} \over 1 - x/2}\,\dd x
=-\,\half\int_{0}^{1}\ln\pars{x}\sum_{n = 0}^{\infty}\pars{x/2}^{n}\,\dd x
=-\,\half\sum_{n = 0}^{\infty}{1 \over 2^{n}}\int_{0}^{1}\ln\pars{x}x^{n}\,\dd x
\\[3mm]&=\left.-\,\half\sum_{n = 0}^{\infty}{1 \over 2^{n}}
\partiald{}{\mu}\int_{0}^{1}x^{n + \mu}\,\dd x\right\vert_{\mu = 0}
=\half\sum_{n = 0}^{\infty}{1 \over 2^{n}\pars{n + 1}^{2}}
\\[3mm]&=\half\
\overbrace{\sum_{n = 0}^{\infty}{1 \over \pars{n + 1}^{2}}}
^{\ds{{\pi^{2} \over 6}}}\
-\ \half\
\overbrace{\sum_{n = 0}^{\infty}{1 - 2^{-n} \over \pars{n + 1}^{2}}}
^{\ds{\ln^{2}\pars{2}}} =
\color{#00f}{\large{\pi^{2} \over 12} - \half\,\ln^{2}\pars{2}}
\end{align}

ADDENDA
Also, it's equivalent to

\begin{align}
\color{#00f}{\large I}&=-\,\half\int_{0}^{1}{\ln\pars{x} \over 1 - x/2}\,\dd x
=-\int_{0}^{1}{\ln\pars{2\bracks{x/2}} \over 1 - x/2}\,{\dd x \over 2}
=-\int_{0}^{1/2}{\ln\pars{2x} \over 1 - x}\,\dd x
\\[3mm]&=-\int_{0}^{1/2}\ln\pars{1 - x}\,\pars{{1 \over 2x}\,2}\,\dd x
=\int_{0}^{1/2}{\rm Li}_{2}'\pars{x}\,\dd x
={\rm Li}_{2}\pars{\half}
\\[3mm]&=\color{#00f}{\large{\pi^{2} \over 12} - \half\,\ln^{2}\pars{2}}
\end{align}

