Evaluate $\int^0_1 \frac{\ln(t)}{1-t^2}dt$ 
Evaluate: $$\int^0_1 \dfrac{\ln(t)}{1-t^2}dt$$

This actually came up while solving another integral. It was suggested that I use a binomial series, but unfortunately I do not understand how to use this. Can anyone help me out?

First consider the operation
\begin{align}
\partial_{n} \, t^{n} = \frac{d}{dn} \, e^{n \ln(t)} = \ln(t) \, e^{n \ln(t)} = t^{n} \, \ln(t).
\end{align}
Now consider the integral, where the operation just presented will be used, 
\begin{align}
I_{n} = \int_{0}^{1} \ln(t) \, t^{n} \, dt = \partial_{n} \, \int_{0}^{1} t^{n} \, dt = \partial_{n} \left[ \frac{t^{n+1}}{n+1} \right]_{0}^{1} = \partial_{n} \left(\frac{1}{n+1}\right) = - \frac{1}{(n+1)^{2}}
\end{align}
Now letting $n \to 2n$ and then summing over $n$ it is seen that:
\begin{align}
\sum_{n=0}^{\infty} I_{2n} = \int_{0}^{1} \frac{\ln(t) \, dt}{1-t^{2}} &= - \sum_{n=0}^{\infty} \frac{1}{(2n+1)^{2}} \\
&= - \left(\sum_{n=0}^{\infty} \frac{1}{(2n+1)^{2}} + \sum_{n=1}^{\infty} \frac{1}{(2n)^{2}} \right) + \frac{1}{4} \sum_{n=1}^{\infty} \frac{1}{n^{2}} \\
&= - \sum_{n=1}^{\infty} \frac{1}{n^{2}} + \frac{1}{4} \sum_{n=1}^{\infty} \frac{1}{n^{2}}  = - \zeta(2) + \frac{1}{4} \, \zeta(2) \\
&= - \frac{3}{4} \zeta(2) = - \frac{\pi^{2}}{8}.
\end{align}
The integral desired is:
\begin{align}
\int_{0}^{1} \frac{\ln(t) \, dt}{1-t^{2}} = - \frac{\pi^{2}}{8}.
\end{align}
