Calculate $I=\int_0^{1}\frac{1+x}{x^2+x+1}\log\left({\frac{x}{1-x}}\right)\,\mathrm dx$ without using complex analysis 
Calculate $$I=\int_0^{1}\frac{1+x}{x^2+x+1}\log\left({\frac{x}{1-x}}\right)\,\mathrm dx$$
  without using complex analysis.

How to calculate without using the residue theorem?
The correct answer is
$$-\frac{1}{8}\ln^23+\frac{\pi^2}{72}-\frac{\pi\sqrt{3}}{36}\ln3$$

This integral is solved using complex analysis in the French book Gilbert Demengel "Balades sur les chemins du champs complexe"page 228 (exemple 4.53.)
The answer in the book $$\frac{1}{8}\ln^23-\frac{\pi^2}{72}-\frac{\pi\sqrt{3}}{36}\ln3$$ is wrong.
 A: \begin{align}
I=&\int_0^{1}\frac{(1+x)\ln{\frac{x}{1-x}}}{x^2+x+1}dx\\
=&\ \frac12\int_0^1 \frac{\ln{\frac{x}{1-x}}}{x^2+x+1}
 + \frac{(2x+1)\ln x}{x^2+x+1}-\frac{(2x+1)\ln (1-x)}{x^2+x+1} \ dx\\
=&\ \frac12(I_1+I_2-I_3)
\end{align}
where
\begin{align}
I_1&=\int_0^1 \frac{\ln{\frac{x}{1-x}}}{x^2+x+1}\overset{\frac {\sqrt3x}{1-x}\to x}{dx}
=\frac1{2\sqrt3}\int_0^\infty \frac{-\ln3}{x^2+\sqrt3x+1}dx
=-\frac{\pi\ln3}{6\sqrt3}\\
I_2&=\int_0^1 \frac{(2x+1)\ln x}{x^2+x+1}{dx}
= 2\Re \text{Li}_2\left(e^{i\frac{2\pi}3}\right)=-\frac{\pi^2}9\\
 I_3&=\int_0^1 \frac{(2x+1)\ln (1-x)}{x^2+x+1}\overset{x\to 1-x}{dx}
= \int_0^1 \frac{(1-\frac23x)\ln x}{\frac13x^2-x+1}{dx}\\
&=-2\Re \text{Li}_2\left(\frac{1+i/\sqrt3}2\right)=\frac14\ln^23-\frac{5\pi^2}{36}
\end{align}
The identity $\Re\text{Li}_2(\frac{1+ia}2)=\frac{\pi^2}{12}-\frac12(\tan^{-1}a)^2-\frac18\ln^2\frac{1+a^2}4$ is used in evaluating $I_3$ and $\Re\text{Li}_2(e^{ia})=-\frac{\pi^2}{12}+\frac14(\pi-a)^2 $ in $I_2$. Substitute $I_1$, $I_2$ and $I_3$ into $I$ to obtain
$$I=\frac{\pi^2}{72}-\frac{\pi}{12\sqrt3}\ln3  -\frac{1}{8}\ln^23$$
