How to evaluate $\int_0^1\frac{\ln(1-2t+2t^2)}{t}dt$?

We have, observing that if $t\in\left(0,1\right)\Rightarrow2t\left(1-t\right)<1$, $$\int_{0}^{1}\frac{\log\left(1-2t+2t^{2}\right)}{t}dt=-\sum_{k\geq1}\frac{2^{k}}{k}\int_{0}^{1}t^{k-1}\left(1-t\right)^{k}dt= $$ $$=-\sum_{k\geq1}2^{k}\frac{\left(k-1\right)!^{2}}{\left(2k\right)!}=-\frac{\pi^{2}}{8} $$ since $$\sum_{k\geq1}2^{2k-1}\frac{\left(k-1\right)!^{2}}{\left(2k\right)!}x^{2k}=\arcsin^{2}\left(x\right),\left|x\right|<1.$$

Integration by parts gives: $$ \int_{0}^{1}\frac{\log(t^2+(t-1)^2)}{t}\,dt = \int_{0}^{1}\frac{2t+2(t-1)}{t^2+(t+1)^2}\,\log(t)\,dt\\=-\int_{0}^{1}\left(\frac{1}{t-\frac{1+i}{2}}+\frac{1}{t-\frac{1-i}{2}}\right)\log(t)\,dt\tag{1}$$ but: $$ \int_{0}^{1}\frac{\log(t)}{t-a}\,dt = \text{Li}_2\left(\frac{1}{a}\right) \tag{2}$$ hence the previous integral is given by $-2\cdot\text{Re}\left[\text{Li}_2(1-i)\right]$, but that can be computed from the reflection formula for $\text{Li}_2$: $$ \text{Li}_2(z)+\text{Li}_2(1-z)=\frac{\pi^2}{6}-\log(z)\log(1-z).\tag{3}$$

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We may also notice that through the substitution $u=2t-1$ the last integral in $(1)$ becomes: $$ I=-\int_{-1}^{1}\frac{2u}{u^2+1}\log\left(\frac{1+u}{2}\right)\,du=-\int_{-1}^{1}\frac{2u}{u^2+1}\log\left(1+u\right)\,du\\=-\int_{-1}^{1}\frac{u}{u^2+1}\log\left(\frac{1+u}{1-u}\right)\,du\tag{4}$$ since the integral of an odd, integrable function over a symmetric interval is zero.

But putting $u=\tanh z$, then $z=-\log v$, we are left with: $$ I = -4\int_{0}^{+\infty}\frac{z\sinh(z)}{\cosh(z)\cosh(2z)}\,dz = 4 \int_{0}^{1}\frac{2v(1-v^2)}{(1+v^2)(1+v^4)}\log(v)\,dv\tag{5}$$ that is: $$ I = 4\int_{0}^{1}\left(\frac{2v}{1+v^2}-\frac{2v^3}{1+v^4}\right)\log(v)\,dv = 3 \int_{0}^{1}\frac{2v}{1+v^2}\log(v)\,dv\tag{6}$$ and by expanding $\frac{2v}{1+v^2}$ as a Taylor series:

$$ I = 3 \sum_{n\geq 0}\left(-\frac{1}{2(n+1)^2}\right) = \frac{3}{2}\sum_{n\geq 1}\frac{(-1)^n}{n^2}=\color{red}{-\frac{\pi^2}{8}}\tag{7}$$

as wanted, avoiding the $\text{Li}_2$ reflection formula (but, in facts, proving it).

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It is interesting to remark that the last approach gives nice rational approximations of $\pi^2$.

For instance, in the same spirit we have: $$ \int_{0}^{1}\left(\frac{v}{1+v^2}+\frac{v^3}{1+v^4}-\frac{v^5}{1+v^6}+\frac{2v^7}{1+v^8}\right)\log(v)\,dv = -\frac{91}{3456}\,\pi^2 $$ where the function between parenthesis is chosen to be $v+v^9+o(v^9)$ in a right neighbourhood of the origin, so that the value of the integral is expected to be close to $$\int_{0}^{1}(v+v^9)\log v\,dv = -\frac{13}{50}$$ and $$ \pi^2\approx\frac{12^3}{7\cdot 5^2} $$ follows.