In the context of the residue theorem, I have this integral to evaluate. The function is even, and $|\int_0^\pi\frac{R^2e^{2i\theta}iRe^{i\theta}}{R^4e^{4i\theta}+6R^2e^{2i\theta} + 13}d\theta| \leq \int_0^\pi2\frac{R^3}{R^4}d\theta \to 0$, so the problem is to find the residues in the upper halfplane.
$\int_0^\infty\frac{x^2}{x^4 + 6x^2 + 13}dx = \frac12\int_{-\infty}^\infty\frac{x^2}{x^4 + 6x^2 + 13}dx = \pi i\sum_{\{\Im z > 0\}}$res$(\frac{x^2}{x^4 + 6x^2 + 13})$
There are two residues to calculate:
$z = \sqrt{-3 + 2i}$: $\frac{\sqrt{-3 + 2i}}{4(-3+ 2i) + 12} = -\frac i8\sqrt{-3 + 2i}$
$z = \sqrt{-3 - 2i}: \frac i8\sqrt{-3 - 2i}$
(Wolfram Alpha if you don't want to trust me)
Giving me overall for the integral:
$\frac\pi8 (\sqrt{-3 + 2i} - \sqrt{-3 - 2i}) = $1.427346... i
But the answer is clearly not meant to be imaginary.
Let us try to avoid useless computations: $x^4+6x^2+13=(x^2+\alpha)(x^2+\beta)$ for a couple of conjugated complex numbers $\alpha,\beta$ with positive real part and such that $\alpha\beta=13$ and $\alpha+\beta=6$.
By partial fraction decomposition we have
$$ \int_{0}^{+\infty}\frac{x^2}{(x^2+\alpha)(x^2+\beta)}\,dx = \frac{1}{\beta-\alpha}\int_{0}^{+\infty}\left(\frac{\beta}{x^2+\beta}-\frac{\alpha}{x^2+\alpha}\right)\,dx = \frac{\pi}{2\left(\sqrt{\beta}+\sqrt{\alpha}\right)}$$ and $$\left(\sqrt{\alpha}+\sqrt{\beta}\right)^2 = \alpha+\beta+2\sqrt{\alpha\beta} = 6+2\sqrt{13} $$ hence the wanted integral equals $\frac{\pi}{2\sqrt{6+2\sqrt{13}}}$. Similarly $$ \int_{0}^{+\infty}\frac{x^2\,dx}{x^4+Ax^2+B} = \frac{\pi}{2\sqrt{A+2\sqrt{B}}} $$
for any $A,B>0$. Lazy is good.