Stuck at this definite integration: $\int_ {0} ^ {\infty} \frac {\log(x)} {x^2 + 2x + 4} \, \mathrm dx$ 
If $$\int_ {0} ^ {\infty} \frac {\log(x)} {x^2 + 2x + 4} \, \mathrm dx \ $$ is equal
to $\pi \ln p/ \sqrt {q} \ $, where $p$ and $q$ are coprimes, then
what is the value of $p + q$?

Ok, so I am stuck with this problem. It was in my exam. The question was a very long one. After solving and simplifying, I was stuck with this definite integration.
The options were $27$ and $29$. I had no idea how to proceed further.
I tried substituting various things, such as $x+1$ to $\tan A$ etc., but nothing seems to work. I hope someone can help shed some light on the problem.
If you want the full question, comment and I will post it.

The hidden trick is that $\int_{0}^{+\infty}\frac{\log(x)}{q(x)}\,dx = 0$ for any quadratic and palindromic polynomial $q(x)$, by the substitution $x\mapsto\frac{1}{x}$. So, if in the original integral we set $x=2y$ we get
$$ \mathcal{I}=\int_{0}^{+\infty}\frac{\log(x)\,dx}{x^2+2x+4} = 2\int_{0}^{+\infty}\frac{\log(2)+\log(y)}{4y^2+4y+4}\,dy = 2\log(2) \int_{0}^{+\infty}\frac{dy}{(2y+1)^2+3} $$
and $\mathcal{I}=\frac{\pi\log(2)}{3\sqrt{3}}$ is a straightforward consequence.
