Calculation of a hard integral $\int^{1/2}_0\frac{1}{x^2-x+1} \ln\frac{1+x}{1-x}\,dx $ Calculate
$$\int^{1/2}_0\frac{1}{x^2-x+1} \cdot \ln\frac{1+x}{1-x}\,dx $$
I tried to subsitute $x=1/2-t$ so $dx=-dt$ but I just complicated more my problem.

Mathematica gives:
$$\frac{\pi  \log \left(\frac{9}{8}\right)-3 i \left(\text{Li}_2\left(-\frac{1}{2}
   (-1)^{2/3}\right)+\text{Li}_2\left(\frac{1}{1-\sqrt[3]{-1}}\right)-\text{Li}_2\left(\frac{1}{1+\sqrt[3]{-1}}\right)+\text{Li}_2\left(\frac{1}{1-(-1)^{2/3}}\right)-\text{Li}
   _2\left(\frac{1}{1+(-1)^{2/3}}\right)+\text{Li}_2\left(\frac{3}{4}-\frac{i
   \sqrt{3}}{4}\right)-\text{Li}_2\left(\frac{1}{4}+\frac{i
   \sqrt{3}}{4}\right)-\text{Li}_2\left(\frac{3}{4}+\frac{i \sqrt{3}}{4}\right)\right)}{3
   \sqrt{3}}$$
which suggests you have a lot of work ahead of you if you want to do this by hand.
