[gobrewers14]

The first integral can be solved replacing the integrand with a series of sort. Notice that the expression inside the logarithm has zeros at

    $\alpha = e^{\pm ix}$

So we can rewrite the function we're integrating as

    $log((\alpha - e^{ix})(\alpha - e^{-ix}))$

which is just

    $2log(\alpha) + log(1 - \frac{e^{ix}}{\alpha}) + log(1 - \frac{e^{-ix}}{\alpha})$

Using

    $log(1 - x) = -\sum_{n=1}^{\infty} \frac{x^n}{n}$

We get

    $2log(\alpha) - \sum\frac{e^{inx}}{n\alpha^{n}}-\sum\frac{e^{-inx}}{n\alpha^{n}}$

which is just

   $2log(\alpha) -2\sum \frac{cos(nx)}{n\alpha^n}$

The integral of the second half of this involves a $sin(nx)$ term which will evaluate to zero for all values of \alpha at 0 and \pi.

Leaving just the integral of $2log(\alpha)$ which is just $2\pi log(\alpha)$