I note that Mathematica could yield the identity $$\int_0^1\frac{\log(1+x^2(x-1)/2)}{x^2(x-1)}dx=\frac{\pi(\pi-4)-12\log^22+24\log2}{16}.\tag{1}\label{1}$$ But I don't know how Mathematica got this.
Question. How to prove \eqref{1} manually?
Your comments are welcome!
With Mathematica, one can actually find an antiderivative $F=G+H$ of the function $f$, where $$f(x):=\frac{\ln(1+x^2(x-1) /2)}{x^2(x-1)},$$ $$G(x):=-\text{Li}_2\left(\frac{1-x}{2}\right)-\frac{1}{2} \text{Li}_2\left(-(x-1)^2\right)+\text{Li}_2(-x)+\text{Li}_2\left(\left(\frac{1}{2}-\frac{i}{2}\right) x\right)+\text{Li}_2\left(\left(\frac{1}{2}+\frac{i}{2}\right) x\right),$$ \begin{gather} H(x):=\frac{\left(x \ln \left(\frac{x-1}{x}\right)+1\right) \ln \left(\frac{1}{2} (x-1) x^2+1\right)}{x}+\ln (x+1) \\ +\ln (x) \left(\ln \left(1-\left(\frac{1}{2}+\frac{i}{2}\right) x\right)+\ln \left(1-\left(\frac{1}{2}-\frac{i}{2}\right) x\right)+\ln (x+1)\right) \\ -\ln (x-1) \left(\ln (i x+(1-i))+\ln \left(\frac{1}{2} ((1+i)-i x) (x+1)\right)\right) \\ -\frac{1}{2} \ln ((x-2) x+2)+\tan ^{-1}(1-x). \end{gather} This can be verified by differentiation, since $\text{Li}'_2(x)=-\ln(1-x)/x$.
The desired result now follows because $$G(0)=\frac{\ln ^2(2)}{2}-\frac{\pi ^2}{24},\quad G(1)=\frac{1}{48} \left(\pi ^2-12 \ln ^2(2)\right),$$ $$\lim_{x\to0}H(x)=(\pi - \ln4)/4,\quad \lim_{x\to1}H(x)=\ln2.$$
The indefinite integral of $f$ can be found "more manually" as follows. Integrate $f$ by parts, to get an integrand of the form $R(x)+R_0(x)\ln x+R_1(x)\ln(1-x)$ instead of $f(x)=\dfrac{\ln(1+x^2(x-1) /2)}{\cdots}$, where $R(x),R_0(x),R_1(x)$ are certain rational expressions. Use then partial fraction decomposition (involving certain complex constants) to reduce the integrand to ones of the form $\frac{\ln u}{u+a}$, and note that $$\int \frac{\ln u}{u+a}\,du= \text{Li}_2\left(-\frac{u}{a}\right)+\ln(u) \ln\left(\frac{a+u}{a}\right)+C.$$
Being lazy, I also computed the antiderivative.
The problem is that Mathematica does not evaluate the result at the bounds since it faces indeterminate forms.
However, series expansions give $$I(1)=\left(\frac{1}{48} \pi (43 \pi -24)+\frac{\log ^2(2)}{4}+\log (2)+i \pi \log (2)\right)+\frac{x-1}{2}+O\left((x-1)^2\right)$$ $$I(0)=\frac{1}{12} \left(-3 \pi +10 \pi ^2+6 \log (2) (-1+2 i \pi +2\log (2))\right)+\frac{x}{2}+O\left(x^2\right)$$ and then the result.
Edit
Using @Iosif Pinelis's suggestion at the end of his answer, integrating by parts lead to
$$\int\frac{\log(1+x^2(x-1)/2)}{x^2(x-1)}dx=\left(\frac{1}{x}+\log (1-x)-\log (x)\right) \log\left(\frac{1}{2} (x-1) x^2+1\right)-$$ $$\int\frac{x (3 x-2) \left(\frac{1}{x}+\log (1-x)-\log (x)\right)}{(x+1) \left(x^2-2 x+2\right)}\,dx$$ Using the bounds, the first term is $0$ $$I_1=\int_0^1\frac{3 x-2}{(x+1) \left(x^2-2 x+2\right)}\,dx=\int_0^1 \left(\frac{x}{x^2-2 x+2}-\frac{1}{x+1}\right)\,dx$$ $$I_1=\frac{1}{4} (\pi -6 \log (2))$$ For the remaining $$\frac{x(3 x-2)}{(x+1) \left(x^2-2 x+2\right)}=\frac{1}{x-(1-i)}+\frac{1}{x+1}+\frac{1}{x-(1+i)}$$ leads to simple integrals and $$\int\frac{x (3 x-2) \left(\log (1-x)-\log (x)\right)}{(x+1) \left(x^2-2 x+2\right)}\,dx=\frac{3 \log ^2(2)}{4}-\frac{\pi ^2}{16}$$ and then the result.
I suppose that this is the simplest manual approach.
