Earlier today I saw this integral around here and gave it a try without success, unfortunately it got taken down so it didn't receive to much attention, but I think it's a nice integral (although it seems quite hard) and finding a closed form it's worth trying.
Evaluate $$I=\int_0^1 \frac{\ln(1+x)}{1+x^3}dx$$
I tried to work with it's sister integral:$$J=\int_0^1 \frac{\ln(1-x)}{1+x^3}dx\Rightarrow I-J=-\int_0^1 \frac{\ln\left(\frac{1-x}{1+x}\right)}{1+x^3}dx$$
Via the substitution $\frac{1-x}{1+x}=t$ we get:
$$I-J=-\int_0^1 \frac{(1+t)\ln t}{1+3t^2}dt=-\sum_{n=0}^\infty (-3)^n \int_0^1 t^{2n}\ln t dt -\sum_{n=0}^\infty (-3)^n \int_0^1 t^{2n+1}\ln t dt $$
$$\int_0^1 x^k dx=\frac{1}{k+1}\overset{\frac{d}{dk}}\Rightarrow \int_0^1 x^k\ln x dx=-\frac{1}{(k+1)^2}$$
$$\Rightarrow I-J=\sum_{n=0}^\infty \frac{(-3)^n }{(2n+1)^2}+\sum_{n=0}^\infty \frac{(-3)^n }{(2n+2)^2}=\sum_{n=0}^\infty \frac{(-3)^n }{(2n+1)^2}+\frac{\operatorname{Li_2 (-3)}}{12}$$
But the first sum is quite ugly looking, so I don't think it's a great approach to the integral and I'm struggling for $I+J$ too.
I remember that OP tried to do partial fractions such as:
$$I=\frac13 \int_0^1 \frac{\ln(1+x)}{1+x}dx-\frac13 \int_0^1 \frac{(x-2)\ln(1+x)}{x^2-x+1}dx$$
And he applied Feynman's trick for the second integral, yet the computation are unbearable and it didn't even spit out the correct result (hopefully there can be a nice closed form and I would like to see one).
Best Answer
The 'sister integral' approach works for quite a few other integrals, but I do not know how to proceed in this particular case as well (note, by the way, that you have used the geometric series outside its radius of convergence in your calculations), so here's a sketch of the somewhat laborious brute-force method:
After the first partial fraction decomposition we have $I = \frac{1}{6} \ln^2(2) + \frac{1}{3} K$, where $$ K = \int \limits_0^1 \frac{(2-x) \ln(1+x)}{1 - x + x^2} \, \mathrm{d} x \, .$$ Now we introduce the sixth root of unity $\alpha \equiv \mathrm{e}^{\mathrm{i} \pi/3} = \frac{1 + \sqrt{3} \mathrm{i}}{2}$. It has the useful properties $\overline{\alpha} = 1- \alpha = - \alpha^2$, $\frac{\alpha}{1+\alpha} = \frac{\mathrm{i} \overline{\alpha}}{\sqrt{3}}$, $\frac{\overline{\alpha}}{1+\alpha} = - \frac{\mathrm{i}}{\sqrt{3}}$ and appears when doing partial fractions once more: $$ \frac{2 - x}{1 - x + x^2} = \frac{- \alpha}{x - \alpha} + \frac{-\overline{\alpha}}{x - \overline{\alpha}} = 2 \operatorname{Re} \left[\frac{- \alpha}{x - \alpha}\right] \, , \, x \in \mathbb{R} \, .$$ Therefore, \begin{align} K &= 2 \operatorname{Re} \left[\alpha \int \limits_0^1 \frac{- \ln(1+x)}{x - \alpha} \, \mathrm{d} x \right] \stackrel{t = x - \alpha}{=} 2 \operatorname{Re} \left[\alpha \int \limits_{-\alpha}^{\overline{\alpha}} \frac{- \ln(1+\alpha) - \ln \left(1 + \frac{t}{1+\alpha}\right) }{t} \, \mathrm{d} t \right] \\ &\hspace{-8pt}\stackrel{s = \frac{-t}{1+\alpha}}{=} 2 \operatorname{Re} \left[\alpha \ln(1+\alpha) \left[\ln(-\alpha) - \ln(\overline{\alpha})\right] + \alpha \int \limits_{\frac{\alpha}{1 + \alpha}}^{-\frac{\overline{\alpha}}{1+\alpha}} \frac{- \ln(1-s)}{s} \, \mathrm{d} s \right] \\ &= \frac{\pi^2}{18} + \frac{\pi \ln(3)}{2 \sqrt{3}} + \operatorname{Re} \left[\operatorname{Li}_2 \left(\frac{\mathrm{i}}{\sqrt{3}}\right) - \operatorname{Li}_2 \left(\frac{\mathrm{i \overline{\alpha}}}{\sqrt{3}}\right)\right] - \sqrt{3} \operatorname{Im} \left[\operatorname{Li}_2 \left(\frac{\mathrm{i}}{\sqrt{3}}\right) - \operatorname{Li}_2 \left(\frac{\mathrm{i \overline{\alpha}}}{\sqrt{3}}\right)\right] \, . \end{align} The dilogarithm values can now be simplified using the various functional equations. We obtain \begin{align} \operatorname{Re} \left[\operatorname{Li}_2 \left(\frac{\mathrm{i}}{\sqrt{3}}\right) \right] &= \frac{1}{2} \operatorname{Li}_2 \left(\frac{1}{3}\right) - \frac{\pi^2}{24} + \frac{1}{8} \ln^2(3) \\ \operatorname{Re} \left[\operatorname{Li}_2 \left(\frac{\mathrm{i \overline{\alpha}}}{\sqrt{3}}\right) \right] &= \frac{5 \pi^2}{72} - \frac{1}{8} \ln^2 (3) \, . \end{align} The imaginary parts are a bit harder to compute (related questions are found here and here), but a reasonably nice expression in terms of the trigamma function can be derived: $$ \operatorname{Im} \left[\operatorname{Li}_2 \left(\frac{\mathrm{i}}{\sqrt{3}}\right) - \operatorname{Li}_2 \left(\frac{\mathrm{i \overline{\alpha}}}{\sqrt{3}}\right)\right] = - \frac{\pi^2}{18 \sqrt{3}} + \frac{\operatorname{\psi}_1 \left(\frac{1}{3}\right)}{12 \sqrt{3}} \, .$$ Thus we arrive at $$ K = \frac{1}{4} \ln^2 (3) + \frac{\pi \ln(3)}{2 \sqrt{3}} + \frac{1}{2} \operatorname{Li}_2 \left(\frac{1}{3}\right) - \frac{1}{12} \operatorname{\psi}_1 \left(\frac{1}{3}\right) $$ and $$ \boxed{I = \frac{1}{6} \ln^2 (2) + \frac{1}{12} \ln^2 (3) + \frac{\pi \ln(3)}{6 \sqrt{3}} + \frac{1}{6} \operatorname{Li}_2 \left(\frac{1}{3}\right) - \frac{1}{36}\operatorname{\psi}_1 \left(\frac{1}{3}\right)} \, .$$ It is of course up to you whether you consider this a nice result, but I have no idea how to simplify it any further.