We have: $$\int_{0}^{1}\frac{\log{\left(t^4-t^2+1 \right)}}{t-1} \,dt=\frac{11}{72}\pi^{2}-\frac{\log^{2} \left(2+\sqrt{3} \right)}{2}\tag{1}.$$ Mathematica and Wolfram Alpha are unable to deal with it (see Wolfram Alpha response). The proof I have follows from the dilogarithm function. As suggested by @J.G here it is: \begin{equation}\mathrm{Li}_2(z)=-\int_{0}^{1}\frac{\log{(1-zy)}}{y}dy,\tag{2} \end{equation} and Euler's formula: \begin{equation}\mathrm{Li}_2(z)+\mathrm{Li}_2(1-z)=\frac{\pi^2}{6}-\log{(1-z)}\log{(z)}.\tag{3} \end{equation} Combining $(2)$ and $(3)$: \begin{align*} \mathrm{Li}_2(z)+\mathrm{Li}_2(1-z)=-\int_{0}^{1}\frac{\log{((1-tz)(t(z-1)+1))}}{t} dt \\=\frac{\pi^2}{6}-\log{(1-z)}\log{(z)},\tag{4} \end{align*} Then if $z_{1}=\frac{1}{2}-\frac{2+\sqrt{3}}{2}i$ and $z_{2}=\frac{1}{2}-\frac{2-\sqrt{3}}{2}i$, we deduce from $(2)$: \begin{align*}i)\hspace{.5cm}\mathrm{Re}\left(\mathrm{Li}_2(z_{1})+\mathrm{Li}_2(z_{2})\right)=\mathrm{Re}\left(-\int_{0}^{1}\frac{\log{(1-z_{1}t)}}{t}dt-\int_{0}^{1}\frac{\log{(1-z_{2}t)}}{t}dt\right)=\\ -\frac{1}{2}\left( \int_{0}^{1}\frac{\log{(t^2(2+\sqrt{3})-t+1)}}{t}dt+ \int_{0}^{1}\frac{\log{(t^2(2-\sqrt{3})-t+1)}}{t}dt\right)= -\frac{1}{2}(I_{1}+I_{2}). \end{align*} In other hand with $(4)$ we obtain: \begin{align*}I_{1}=\int_{0}^{1}\frac{\log{((1-tz_{1})(t(z_{1}-1)+1))}}{t} dt=\int_{0}^{1}\frac{\log{(t^2(2+\sqrt{3})-t+1)}}{t}dt =-\frac{\pi^2}{6}+\log{(1-z_{1})}\log{(z_{1})}\\=\frac{\pi^2}{144}+\frac{\log^2{(2+\sqrt{3})}}{4},\end{align*} and \begin{align*}I_{2}=\int_{0}^{1}\frac{\log{((1-tz_{2})(t(z_{2}-1)+1))}}{t} dt=\int_{0}^{1}\frac{\log{(t^2(2-\sqrt{3})-t+1)}}{t}dt=-\frac{\pi^2}{6}+\log{(1-z_{2})}\log{(z_{2})}\\=-\frac{23\pi^2}{144}+\frac{\log^2{(2-\sqrt{3})}}{4}=-\frac{23\pi^2}{144}+\frac{\log^2{(2+\sqrt{3})}}{4}.\end{align*} Then replacing $I_{1}$ and $I_{2}$ in $i)$ we have: $$\mathrm{Re}\left(\mathrm{Li}_2(z_{1})+\mathrm{Li}_2(z_{2})\right)=\frac{11}{144}\pi^{2}-\frac{\log^{2}(2+\sqrt{3})}{4}.\tag{5}$$ Then observe that: $$I_{1}+I_{2}=\int_{0}^{1}\frac{\log{(-(y^2(\sqrt{3}+2)-y+1)(y^2(\sqrt{3}-2)+y-1))}}{y}dy,$$ since $-(y^2(\sqrt{3}+2)-y+1)(y^2(\sqrt{3}-2)+y-1)=y^4-4y^3+5y^2-2y+1$, then: $$I_{1}+I_{2}=\int_{0}^{1}\frac{\log{(y^4-4y^3+5y^2-2y+1)}}{y}dy\overset{y=1-t}=\int_{0}^{1}\frac{\log{(t^4-t^2+1)}}{t-1}dt,$$ and the result follows from $(i)$ and $(5)$.
Question: Can we prove this integral using known series expansions or in an alternative way?
A proof without using the dilogarithm "function" is so appreciated.
$$I=\int_0^1 \frac{\ln(1-x^2+x^4)}{1-x}dx=\int_0^1\ln(1-x^2+x^4)\left(\frac{x}{1-x^2}+\frac{1}{1-x^2}\right)dx$$
$$=-\frac{\pi^2}{36}+\frac{\ln^2(2+\sqrt 3)}{2}-\frac{\pi^2}{8}=\boxed{\frac{\ln^2(2+\sqrt 3)}{2}-\frac{11\pi^2}{72}}$$
$$\int_0^1 \frac{x\ln(1-x^2+x^4)}{1-x^2}dx\overset{1-x^2\to x}=\frac12\int_0^1 \frac{\ln(1-x+x^2)}{x}dx$$ $$=\frac12\underbrace{\int_0^1\frac{\ln(1+x^3)}{x}dx}_{x^3\to x}-\frac12\int_0^1\frac{\ln(1+x)}{x}dx=-\frac13\int_0^1\frac{\ln(1+x)}{x}dx=-\frac{\pi^2}{36}$$
$$\mathcal K=\int_0^1 \frac{\ln(1-x^2+x^4)}{1-x^2}dx$$
$$\small \overset{\large x\to\frac{1-x}{1+x}}=\frac12\underbrace{\int_0^1\frac{\ln(1+14x^2+x^4)}{x}dx}_{x^2\to x}-2\int_0^1\frac{\ln(1+x)}{x}dx=\frac14\mathcal J(\operatorname{arccosh} 7)-\mathcal J(0)$$
$$\small \mathcal J(t)=\int_0^1 \frac{\ln(1+2x\cosh t +x^2)}{x}dx \Rightarrow \mathcal J '(t)=2\int_0^1 \frac{\sinh t}{(x+\cosh t)^2-\sinh^2 t}dx=t$$
$$\Rightarrow \mathcal K= \frac14\int_0^{\operatorname{arccosh} 7} t\, dt -\frac34\mathcal J(0)= \frac{\ln^2(2+\sqrt 3)}{2}-\frac{\pi^2}{8}$$