How to compute $\int_0^\infty \frac{\log(2+x^2)}{4+x^2}\,\mathrm dx$ Evaluate the integral
$$\int_0^\infty \frac{\log(2+x^2)}{4+x^2}dx$$
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I started by stating that the integral from 0 to infinity should be the same as half the integral from $-\infty$ to $\infty$, that is:
$$\int_0^\infty \frac{\log(2+x^2)}{4+x^2}dx = \frac{1}{2}\int_{-\infty}^\infty \frac{\log(2+x^2)}{4+x^2}dx$$
and then by stating that this must be equal to 
$$\pi i \cdot \sum(\text{residues in upper plane})$$
noting that there is a singularity at $z=2i$ in the upper half plane, and an "issue" (I don't know if it's strictly a singularity) with the log function at $z=\sqrt{2}i$.
The residue at $z=2i$ is easily dealt with, but when I try to deal with the log issue, I can't make any headway. I decided to make my branch cut between $z=\sqrt{2}i$ and $z=-\sqrt{2}i$, and form a contour that goes around this cut and the point, but it doesn't seem to be working out for me. 
Suggestions would be appreciated! 

Without residues.
$$ \frac{\log(2+x^2)}{4+x^2}=\frac{\log(x+i\sqrt2)+\log(x-i\sqrt2)}{(x+2i)(x-2i)}$$
$$\frac{1}{(x+2i)(x-2i)}=\frac{i}{4}\left(\frac{1}{x+2 i}-\frac{1}{x-2 i}\right)$$ So, we face four integrals looking like
$$\int \frac{\log(x+a)}{x+b}\,dx=\text{Li}_2\left(\frac{x+a}{a-b}\right)+\log (x+a) \log \left(\frac{b+x}{b-a}\right)$$
