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?
 A: 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}
A: 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.
A: Substitute $t\mapsto e^{-t}$:
$$
\begin{align}
\int_1^0\frac{\log(t)}{1-t^2}\,\mathrm{d}t
&=\int_0^\infty\frac{t}{1-e^{-2t}}e^{-t}\,\mathrm{d}t\\
&=\sum_{k=0}^\infty\int_0^\infty te^{-(2k+1)t}\,\mathrm{d}t\\
&=\Gamma(2)\sum_{k=0}^\infty\frac1{(2k+1)^2}\\
&=\frac{\pi^2}8
\end{align}
$$
