The other ways to calculate $\int_0^1\frac{\ln(1-x^2)}{x}dx$ 
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. (>‿◠)✌

Using the dilogarithm $\mathrm{Li}_2\;$ and the particular values for $0,1,-1\;$you get:
$$\int_0^1\frac{\ln(1-x^2)}{x}dx= \int_0^1\frac{\ln(1-x)(1+x)}{x}dx=
\int_0^1\frac{\ln(1+x)}{x}dx + \int_0^1\frac{\ln(1-x)}{x}dx=
-\mathrm{Li}_2(-x)\Big{|}_0^1 - \mathrm{Li}_2(x)\Big{|}_0^1
=\frac{\pi^2}{12}-\frac{\pi^2}{6} = -\frac{\pi^2}{12}$$
