Find integral $\int_0^1 \frac{\ln x}{1 - x^2} \mathrm{d}x$ I have to find definite integration $$\int_0^1 \frac{\ln x}{1 - x^2} \mathrm{d}x$$ 
I tried to subtitute $x = \sin u$  and $x = e^u$ 
but got no result . Please help me in proceeding.

The first step is to turn the one integral into two as follows:
\begin{align}
\int_{0}^{1} \frac{\ln(x)}{1-x^2} \, dx = \frac{1}{2} \, \left[ \int_{0}^{1} \frac{\ln(x)}{1-x} \, dx + \int_{0}^{1} \frac{\ln(x)}{1+x} \, dx \right]
\end{align}
then using the integrals
\begin{align}
\int_{0}^{1} \frac{\ln(x)}{1-x} \, dx &= \left[Li_{2}(1-x) \right]_{0}^{1} = - \zeta(2) = - \frac{\pi^2}{6} \\
\int_{0}^{1} \frac{\ln(x)}{1+x} \, dx &= - \frac{\pi^2}{12}
\end{align}
the desired integral becomes
$$\int_{0}^{1} \frac{\ln(x)}{1-x^2} \, dx = - \frac{\pi^2}{8}.$$
