Integrate $\int\limits_{0}^{1} \frac{\log(x^{2}-2x \cos a+1)}{x} dx$ How do I solve this: $$\int\limits_{0}^{1} \frac{\log(x^{2}-2x \cos{a}+1)}{x} \ dx$$
Integration by parts is the only thing which I could think of, clearly that seems cumbersome. Substitution also doesn't work.

Apply the identity $Li_2(z) + Li_2({1/z})=-\frac{\pi^2}6-\frac12\ln^2(-z)$ in the evaluation below
\begin{align}
\int_{0}^{1} \frac{\ln(x^{2}-2x\cos{a}+1)}{x} \ dx 
=& \int\limits_{0}^{1} \frac{\ln(1-xe^{i a})+ \ln(1-xe^{-ia })}{x} \ dx  \\=&-Li_2(e^{ia })- Li_2(e^{-ia }) \\
=& \frac{\pi^2}6 + \frac12\ln^2(-e^{ia})= \frac{\pi^2}6+\frac12[i (a-\pi)]^2\\
= &-\frac12a^2 +\pi a - \frac{\pi^2}3
\end{align}
