Maybe differentiation can be avoided.

Let

$$A\left( a \right) = \int\limits_0^\pi {\log \left( {1 + a \cos x} \right)dx} $$

Then with $x\mapsto x-\pi$ we get $$A\left( a \right) = \int\limits_0^\pi {\log \left( {\frac{1}{{1 - a\cos x}}} \right)dx} $$

or

$$A\left( a \right) = \frac{1}{2}\int\limits_0^\pi {\log \left( {\frac{{1 + a\cos x}}{{1 - a\cos x}}} \right)dx} $$

Now, use the Weierstrass substitution, $$\tan \left( {\frac{x}{2}} \right) = t$$

to get $$\eqalign{
& dx = \frac{{2dt}}{{1 + {t^2}}} \cr