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Prove that $$\sf I=\int_0^1 \frac{\ln(1+x+x^2)\ln(1-x+x^2)}{x}dx=\frac{\pi}{6\sqrt{3}}\psi_1\left(\frac{1}{3}\right)-\frac{\pi^3}{9\sqrt{3}}-\frac{19}{18}\zeta(3).$$

I have thought about the integral from above after I saw this similar integral and I believe changing the sign to have $\sf 1+x+x^2$ might get us a nice closed form.

So I started using the following formula: $$\sf 2ab=(a+b)^2-a^2-b^2$$ $$\sf \Rightarrow 2I=\int_0^1\frac{\ln^2(1+x^2+x^4)}{x}dx-\int_0^1\frac{\ln^2(1+x+x^2)}{x}dx-\int_0^1\frac{\ln^2(1-x+x^2)}{x}dx$$ Using in the first integral the substitution $\sf x^2\rightarrow x $ gets us: $$\sf \int_0^1\frac{\ln^2(1+x^2+x^4)}{x}dx=\frac12\int_0^1\frac{\ln^2(1+x+x^2)}{x}dx$$ $$\sf \Rightarrow I=-\frac14\int_0^1\frac{\ln^2(1+x+x^2)}{x}dx-\frac12\int_0^1\frac{\ln^2(1-x+x^2)}{x}dx$$ Well, now we only need to find: $$\sf I(a)=\int_0^1\frac{\ln^2(1+ax+x^2)}{x}dx $$ Then set $a=1$ and $a=-1$. Of course I tried to use Feynman's trick: $$\sf I'(a)=2\int_0^1\frac{\ln(1+ax+x^2)}{1+ax+x^2}dx$$ But quickly gave up as it doesn't look promising.

Another way might be to let $\sf x+\frac12=\frac{\sqrt 3}{2}t$ in order to get: $$\sf \int_0^1\frac{\ln^2(1+x+x^2)}{x}dx=\int_\frac{1}{\sqrt 3}^\sqrt 3 \frac{\ln^2\left(\frac34(1+t^2)\right)}{t-\frac{1}{\sqrt 3}}dt$$ But well.. I would appreciate some help!


Update. In the meantime I found something a conjecture: $$\sf \int_0^1\frac{\ln^2(1+x+x^2)}{x}dx=\frac{2\pi}{9\sqrt3}\psi_1\left(\frac13\right)-\frac{4\pi^3}{27\sqrt3}-\frac23\zeta(3)$$

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As shown in the question we have: $$\sf I=-\frac14\int_0^1\frac{\ln^2(1+x+x^2)}{x}dx-\frac12\int_0^1\frac{\ln^2(1-x+x^2)}{x}dx$$ For the first integral we can write: $$\sf (a-b)^2=a^2-b^2-2b(a-b);\ a=\ln(1-x^3),b=\ln(1-x)$$ $$\sf \Rightarrow \int_0^1\frac{\ln^2(1+x+x^2)}{x}dx=\int_0^1 \frac{\left(\ln(1-x^3)-\ln(1-x)\right)^2}{x}dx$$ $$\sf =\color{blue}{\int_0^1 \frac{\ln^2(1-x^3)}{x}dx}-\int_0^1 \frac{\ln^2(1-x)}{x}dx-2\int_0^1 \frac{\ln(1-x)\ln(1+x+x^2)}{x}dx$$ $$\sf \overset{\color{blue}{x^3\to x}}=-\frac23\int_0^1 \frac{\ln^2(1-x)}{x}dx-2\int_0^1 \frac{\ln(1-x)\ln(1+x+x^2)}{x}dx=-\frac43\zeta(3)-2J$$ Note also that: $$\sf \int_0^1 \frac{\ln^2(1-x)}{x}dx=\int_0^1 \frac{\ln^2 x}{1-x}dx=\sum_{n=1}^\infty \int_0^1 x^{n-1}\ln^2 x\, dx=2\sum_{n=1}^\infty \frac{1}{n^3}=2\zeta(3)$$ The latter integral can be found here: $$\sf J=\int_0^1 \frac{\ln(1-x)\ln(1-x+x^2)}{x}dx=-\frac{\pi}{9\sqrt 3}\psi_1\left(\frac13\right)+\frac{2\pi^3}{27\sqrt 3}-\frac13\zeta(3) $$ $$\Rightarrow \boxed{\sf \int_0^1\frac{\ln^2(1+x+x^2)}{x}dx=\frac{2\pi}{9\sqrt3}\psi_1\left(\frac13\right)-\frac{4\pi^3}{27\sqrt3}-\frac23\zeta(3)}$$ The second integral can be found here: $$\boxed{\sf\int_0^1 \frac{\ln^2(1-x+x^2)}{x}dx=-\frac{4\pi}{9\sqrt{3}}\psi_1\left(\frac{1}{3}\right)+\frac{8\pi^3}{27\sqrt{3}}+\frac{22}{9}\zeta(3)}$$ Combining the boxed results yields: $$\boxed{\sf \int_0^1 \frac{\ln(1+x+x^2)\ln(1-x+x^2)}{x}dx=\frac{\pi}{6\sqrt{3}}\psi_1\left(\frac{1}{3}\right)-\frac{\pi^3}{9\sqrt{3}}-\frac{19}{18}\zeta(3)}$$

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