Consider the following attempt to calculate $\int_0^\infty \frac{x^2}{1+x^4} \, dx$:
First we note that $f(x) = \frac{x^2}{1+x^4}$ is even, so $$\int_0^\infty \frac{x^2}{1+x^4} \, dx = \frac{1}{2} \int_{-\infty}^\infty \frac{x^2}{1+x^4} \, dx $$
Next, consider that $$\frac{x^2}{1+x^4} = \frac{x^2}{(x-i)(x+i)(x-1)(x+1)}$$ and so $f(x)$ has 4 simple poles.
Then to find the residues, we are allowed to take \begin{align*} \lim_{x \to i}{(x-i)f(x)} &= \lim_{x \to i} \frac{(x-i)x^2}{(x-i)(x+i)(x-1)(x+1)} = \frac{i^2}{(2i)(i^2-1)} = \frac{-1}{2i^3-2i} \\ \lim_{x \to -i}{(x+i) f(x)} &= \dots \\ \lim_{x \to -1}{(x+1) f(x)} &= \dots \\ \lim_{x \to 1}{(x-1) f(x)} &= \dots \end{align*}
But I'm noticing that when I do this, I am getting complex residues; I don't believe I've ever obtained complex residues before. Although, at the same time, I couldn't see why the residues couldn't be complex. Is this okay or have I made an error somewhere? Thanks in advance.