$\int_0^1 {\frac{{\ln (1 - x)}}{x}}$ without power series Is there a way to calculate $$\int_0^1{ \ln (1 - x)\over x}\;dx$$ without using power series?
 A: A related problem. Using the change of variables $x=1-e^{-t}$ and taking advatage of the fact that 
$$\Gamma(s)\zeta(s) = \int_{0}^{\infty} \frac{t^{s-1}}{e^{t}-1}\,, $$ 
the value of the integral follows
$$ -\int_{0}^{\infty} \frac{t}{e^{t}-1} \,dt = -\zeta(2) = -\frac{\pi^2}{6}   \,.$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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\begin{align}
& \color{#44f}{\int_{0}^{1}{\ln\pars{1 - x} \over x}
\dd x} =
\left.\partiald{}{\mu}\int_{0}^{1}x^{\epsilon\, - 1}
\bracks{\pars{1 - x}^{\mu} - 1}\dd x
\right\vert_{\substack{\epsilon\ =\ 0^{+}\\[1mm] \mu\ =\ 0\,\,}}
\\[5mm] = & \
\partiald{}{\mu}\bracks{{\Gamma\pars{\epsilon}\Gamma\pars{\mu + 1} \over \Gamma\pars{\epsilon + \mu + 1}} - {1 \over \epsilon}}
_{\substack{\epsilon\ =\ 0^{+}\\[1mm] \mu\ =\ 0\,\,}}
\\[5mm] = & \
\partiald{}{\mu}\braces{{1 \over \epsilon}\bracks{{\Gamma\pars{1 + \epsilon}\Gamma\pars{\mu + 1} \over \Gamma\pars{\epsilon + \mu + 1}} - 1}}
_{\substack{\epsilon\ =\ 0^{+}\\[1mm] \mu\ =\ 0\,\,}}
\\[5mm] = & \
\left.-\,\partiald{H_{\mu}}{\mu}
\right\vert_{\mu\ =\ 0} = \bbx{\color{#44f}{-\,{\pi^{2} \over 6}}}
\approx -1.6449 \\ &
\end{align}
A: \begin{align}
&\int_0^1 {\frac{{\ln (1 - x)}}{x}}dx\\
=& \ \frac43 \bigg( \int_0^1 {\frac{{\ln (1 - x)}}{x}}
\overset{x\to \frac{1-x}{1+x}} {dx}  - \frac14\int_0^1 {\frac{{\ln (1 - x)}}{x}}
\overset{x\to (\frac{1-x}{1+x} )^2} {dx}\bigg)\\
=& \ \frac43 \int_0^1 \frac{\ln x}{1-x^2}dx
=\frac43 \int_0^1 \int_0^\infty \frac{-y}{(1+y^2)(1+x^2y^2)}dy\ dx \\
=& -\frac43\int_0^\infty\frac{\tan^{-1}y}{1+y^2}dy
=-\frac43\cdot \frac12 (\tan^{-1}y)^2\bigg|_0^\infty=-\frac{\pi^2}6
\end{align}
