I know that if f(x) is an even function then there's a formula involving the sum of residues to compute the improper integral. But how then would you solve something like $\int_{-\infty}^{\infty} \frac{dx}{x^{2} + 2x + 5}$? Because here f(x) isn't even. I don't really know where to start and can't find relevant examples online or in my textbook.
(Not a homework question - I'm studying for a final).
On
Let $\gamma_R$ be a path that goes from $-R$ to $R$ along a straight line followed from going from $R$ to $-R$ along the semicircle with radius $R$ centered at $0$ and located in the upper half-plane ($R\gg0$). The\begin{align}\int_{\gamma_R}f(z)\,\mathrm dz&=\int_{\gamma_R}\frac1{(z+1)^2+4}\,\mathrm dz\\&=2\pi i\operatorname{Res}_{-1+2i}\left(\frac1{(z+1)^2+4}\right)\\&=\frac{2\pi i}{4i}\\&=\frac\pi2.\end{align}
On
If you want a real-analytic method, you can complete the square on the denominator and then make a simple substitution:
\begin{align} \int_{-\infty}^{\infty} \frac{dx}{x^2+2x+5}&=\int_{-\infty}^{+\infty} \frac{dx}{(x+1)^2+4} \\ \\&\overset{x\mapsto u-1}=\int_{-\infty}^{\infty} \frac{du}{u^2+4} \\ \\ &=\frac{1}{2}\arctan\left(\frac{u}{2}\right)\Big|_{-\infty}^{\infty} \\ \\ &=\frac{\pi}{2} \end{align}
The fact that $f$ is not even is irrelevant. Even-odd symmetries would be important when computing an integral on $[0, \infty)$, for example.
The function $(x^2 + 2x + 5)^{-1} = \big((x + 1)^2 + 4\big)^{-1}$ has a pole in the upper half plane at $-1 + 2i$ with residue $1/4i$.
Thus the integral over a contour $C_R$ consisting of the segment $[-R, R]$ together with a semicircle connecting the endpoints is
$$\int_{C_R} \frac{1}{z^2 + 2z + 5} \, dz = 2\pi i \cdot \frac 1 {4i} = \frac {\pi}{2}.$$
The decay over the semicircle is easy to handle, and so the original integral is $\pi/2$.
This can also be done easily with real variable techniques, because after an easy change of variables we have $\frac 1 2 \int_{-\infty}^{\infty} \frac 1 {x^2 + 1} \, dx$.