How to evaluate $\int_0^1\frac{\ln(1-2t+2t^2)}{t}dt$? The question starts with:
$$\int_0^1\frac{-2t^2+t}{-t^2+t}\ln(1-2t+2t^2)dt\text{ = ?}$$  

My attempt is as follows:
$$\int_0^1\frac{-2t^2+t}{-t^2+t}\ln(1-2t+2t^2)dt$$
$$=2\int_0^1\ln(1-2t+2t^2)dt+\int_0^1\frac{-t}{-t^2+t}\ln(1-2t+2t^2)dt$$
$$=-4+\pi-\frac{1}{2}\int_0^1\frac{\ln(1-2t+2t^2)}{-t^2+t}dt$$
$$=-4+\pi-\frac{1}{2}(\int_0^1\frac{\ln(1-2t+2t^2)}{t}dt+\int_0^1\frac{\ln(1-2t+2t^2)}{1-t}dt)$$
$$=-4+\pi-\int_0^1\frac{\ln(1-2t+2t^2)}{t}dt$$  

The question left with: how can I evaluate:
$$\int_0^1\frac{\ln(1-2t+2t^2)}{t}dt$$
Wolfram Alpha gives me the result: $-\frac{\pi^2}{8}$, but I am not able to obtain this result by hand.

As OP has found
\begin{equation*}
I = \int_0^1\dfrac{\ln(1-2t+2t^2)}{t}\, dt = \dfrac{1}{2}\int_0^1\dfrac{\ln(1-2t+2t^2)}{t(1-t)}\, dt.
\end{equation*}
Via the transformation $s= \dfrac{t}{1-t}$ and a partial integration we get
\begin{equation*}
I = \dfrac{1}{2}\int_0^{\infty}\dfrac{\ln(1+s^2)-2\ln(1+s)}{s}\, ds = \int_0^{\infty}\dfrac{(1-s)\ln(s)}{(s+1)(s^2+1)}\, ds.
\end{equation*}
After integrating 
\begin{equation*}
f(s) = \dfrac{(1-s)\log^2(s)}{(s+1)(s^2+1)}
\end{equation*}
along a keyhole or a branch cut we have
\begin{equation*}
\int_0^{\infty}\dfrac{(1-s)(\ln^2(s)-(\ln(s)+i2\pi)^2)}{(s^2+1)(s+1)}\, ds = 2\pi i\sum_{s=-1,\pm i}{\rm Res} f(s).
\end{equation*}
From this we can extract that $I = -\dfrac{\pi^{2}}{8}.$
