Evaluate $\int_{-\infty}^{\infty}\frac{1}{(x^2+4)^5}dx$ $$\int_{-\infty}^{\infty}\frac{1}{(x^2+4)^5}dx$$
I am trying to use residue.
We first need to find the singularities, $x^2+4=0\iff x=\pm 2i$
Just $2i$ is in the positive part of $i$ so we take the limit
$lim_{z\to 2i}\frac{1}{(z^2+4)^5}$ but the limit is $0$

For any $a>0$ we have
$$ \int_{-\infty}^{+\infty}\frac{dx}{x^2+a}=\frac{\pi}{\sqrt{a}} $$
and by applying $\frac{d^4}{da^4}$ to both sides we get:
$$ 24\int_{-\infty}^{+\infty}\frac{dx}{(x^2+a)^5}=\frac{105 \pi}{16 a^4\sqrt{a}} $$
so by rearranging and evaluating at $a=4$ we get:
$$ \int_{-\infty}^{+\infty}\frac{dx}{(x^2+4)^5} = \color{red}{\frac{35 \pi}{2^{16}}}.$$
