TITLE: The other ways to calculate $\int_0^1\frac{\ln(1-x^2)}{x}dx$
QUESTION [15 upvotes]: Prove that
  $$\int_0^1\frac{\ln(1-x^2)}{x}dx=-\frac{\pi^2}{12}$$
  without using series expansion.


An easy way to calculate the above integral is using series expansion. Here is an example
\begin{align}
\int_0^1\frac{\ln(1-x^2)}{x}dx&=-\int_0^1\frac{1}{x}\sum_{n=0}^\infty\frac{x^{2n}}{n} dx\\
&=-\sum_{n=0}^\infty\frac{1}{n}\int_0^1x^{2n-1}dx\\
&=-\frac{1}{2}\sum_{n=0}^\infty\frac{1}{n^2}\\
&=-\frac{\pi^2}{12}
\end{align}
I am wondering, are there other ways to calculate the integral without using series expansion of its integrand? Any method is welcome. Thank you. (>‿◠)✌

REPLY [3 votes]: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}&\color{#66f}{\large\int_{0}^{1}{\ln\pars{1 - x^{2}} \over x}\,\dd x}
=\half\int_{0}^{1}{\ln\pars{1 - x} \over x}\,\dd x
=\half\int_{0}^{1}{\ln\pars{x} \over 1 - x}\,\dd x
\\[3mm]&=-\,\half\lim_{\mu \to 0}\partiald{}{\mu}\int_{0}^{1}
{1 - x^{\mu} \over 1 - x}\,\dd x
=-\,\half\lim_{\mu \to 0}\partiald{\color{#f00}{\Psi\pars{\mu + 1}}}{\mu}
=-\,\half\,\Psi'\pars{1}
\\[3mm]&=-\,\half\,\color{#f0f}{\zeta\pars{2}}
=-\,\half\,\color{maroon}{\pi^{2} \over 6}=\color{#66f}{\large -\,{\pi^{2} \over 12}}
\end{align}

See $\underline{\color{#f00}{6.3.22}}$,
  $\underline{\color{#f0f}{6.4.2}}$ and
  $\underline{\color{maroon}{23.2.24}}$.