Conjecture $\int_0^1\frac{\ln^2\left(1+x+x^2\right)}x dx\stackrel?=\frac{2\pi}{9\sqrt3}\psi^{(1)}(\tfrac13)-\frac{4\pi^3}{27\sqrt3}-\frac23\zeta(3)$

Solution 1:

1. Replace $\ln(1+x+x^2)$ by $\ln(1-x^3)-\ln(1-x)$.

2. Two of the resulting integrals are easy to compute: $$\int_0^1\frac{\ln^2\left(1-x^3\right)dx}{x}=\frac23\zeta(3),\quad \int_0^1\frac{\ln^2\left(1-x\right)dx}{x}=2\zeta(3).$$

3. Mathematica computes and fullsimplifies the remaining nontrivial integral $\int_0^1\frac{\ln\left(1-x\right)\ln\left(1-x^3\right)dx}{x}$ to a one-line expression containing a sum of two trilogarithms $\operatorname{Li}_3(z_1)+\operatorname{Li}_3(z_2)$.

4. It so happens that $z_1+z_2=1$, hence thanks to Landen's identity the above sum is equal to $$-\operatorname{Li}_3\left(\frac{z_1}{z_1-1}\right)+\zeta\left(3\right)+\text{elementary}.$$

5. Finally, it so happens that $\frac{z_1}{z_1-1}=e^{2\pi i/3}$ and the corresponding trilogarithmic value is known in terms of $\zeta(3)$.

Altogether this should lead to your answer.

Solution 2:

By some simple manipulations the integral gets reduced to calculating

$$\frac{1}{2}\int_1^3 \frac{\log ^2(x)}{x-1} \ dx+\frac{1}{2}\int_1^3\frac{\log ^2(x)}{ (x-1) \sqrt{4 x-3}} \ dx,$$

and both residual integrals are pretty easy to finish at this point.