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!