<s>[INST] 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.
 [/INST] Substitute $t=\frac{1-x}{1+x}$
\begin{align}
&\int^{1/2}_0\frac{1}{x^2-x+1} \ln\frac{1+x}{1-x}\,dx \\
=&-2\int_{\frac13}^1 \frac{\ln t}{1+3t^2}dt
\overset{\sqrt3t\to t}= -\frac2{\sqrt3} \int_{\frac1{\sqrt3}}^{\sqrt3} \frac{\ln t-\ln\sqrt3}{1+t^2}dt\\
=& \frac{\ln 3}{\sqrt3} \int_{\frac1{\sqrt3}}^{\sqrt3} \frac{1}{1+t^2}dt = \frac{\pi\ln3}{6\sqrt3}
\end{align}
Note $\int_{\frac1{\sqrt3}}^{\sqrt3} \frac{\ln t}{1+t^2}dt \overset{t\to\frac1t} =0$
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