How to evaluate this integral? $\int^1_0\frac{(\frac{1}{2}-x)\ln (1-x)}{x^2-x+1}\mathrm{d}x$ $$\int^1_0\frac{\left(\frac{1}{2}-x\right)\ln (1-x)}{x^2-x+1}\mathrm{d}x$$
The indefinite integral didn't seem to be helpful...which include the $\operatorname{Li}(x)$.
Also, I can't set up a parameter to differentiate it.
Any hint will be helpful. Thanks!
 A: First, substitute $y=1-x$:
$$
\Rightarrow \int _0^1 \frac{(y-1/2)\log(y)}{y^2-y+1}\,dy
$$Then use integration by parts with $u=\log(y)$:
$$
=\left. \frac{1}{2}\log(y^2-y+1)\log(y)\right|_{0}^{1} -\frac{1}{2}\int_0^1\frac{\log(y^2-y+1)}{y}\,dy
$$The boundary terms vanish (check this yourself). Note that $y^3+1=(y^2-y+1)(y+1)$, so we have
$$
 -\frac{1}{2}\int_0^1\frac{\log(y^2-y+1)}{y}\,dy =  -\frac{1}{2}\int_0^1\frac{\log((y^3+1)/(y+1))}{y}\,dy
$$
$$
=  \frac{1}{2}\int_0^1\frac{\log(y+1)}{y}\,dy- \frac{1}{2}\int_0^1\frac{\log(y^3+1)}{y}\,dy
$$Now invoke the Maclaurin series for $\log(1+z)$:
$$
= \frac{1}{2} \int _0^1 \sum_{k=0}^{\infty} \frac{(-1)^k y^k}{k+1}\,dy-\frac{1}{2} \int _0^1 \sum_{k=0}^{\infty} \frac{(-1)^k y^{3k+2}}{k+1}\,dy
$$Switch the integral and the sum and use the power rule:
$$
= \frac{1}{2} \sum_{k=0}^{\infty} \int _0^1 \frac{(-1)^k y^k}{k+1}\,dy-\frac{1}{2} \sum_{k=0}^{\infty}\int _0^1  \frac{(-1)^k y^{3k+2}}{k+1}\,dy
$$
$$
= \frac{1}{2} \sum_{k=0}^{\infty} \frac{(-1)^k}{(k+1)^2}-\frac{1}{2\cdot 3} \sum_{k=0}^{\infty} \frac{(-1)^k }{(k+1)^2}
$$
$$
= \frac{1}{3} \sum_{k=0}^{\infty} \frac{(-1)^k}{(k+1)^2}
$$If you buy that $\sum_{k=0}^{\infty} \frac{1}{(k+1)^2}=\frac{\pi^2}{6}$, you can convince yourself the alternating version sums to $\frac{\pi^2}{12}$. This gives the total of $\frac{\pi^2}{36}$.
