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?

The geometric series formula tells you that
$$\frac{1}{1-r}=\sum_{n=0}^\infty r^n$$
if $|r|<1$. Applying this to $r=t^2$ you get that your integral is
$$\int_1^0 \sum_{n=0}^\infty \ln(t) t^{2n} dt.$$
You can interchange the sum and integral, for example using monotone convergence (since the integrands are all negative), so you have
$$\sum_{n=0}^\infty \int_1^0 \ln(t) t^{2n} dt.$$
Each of these integrals can be done using integration by parts with $u=\ln(t)$ and $dv=t^{2n} dt$. They are improper at the endpoint of $0$, but this is no real obstacle, because the log term in each antiderivative is getting multiplied with a monomial, so the log terms in the definite integrals all vanish.
