How do I use residue theorem to evaluate this improper integral to get a good looking solution? The problem is $\int_{0}^{\infty} \frac{\sqrt{x}}{x^2+2x+5}dx$ 
I replace x with z, and did some algebra, but the solution was rather nasty. it contains exponential and arctan such and such. However, the solution given was $\frac{\pi}{2\sqrt{2}}{\sqrt{\sqrt{5}-1}}$. It was pretty neat and I have no idea how I could convert my solution to that form. 
 A: Suppose we seek to evaluate
$$J = \int_0^\infty \frac{\sqrt{x}}{x^2+2x+5} dx$$
using a keyhole contour and the function
$$f(z) = \frac{\exp(1/2\log(z))}{z^2+2z+5}$$
with the branch cut of the logarithm on the positive real axis and the
argument from $0$ to $2\pi.$

We obtain that
$$J (1- e^{\pi i})
= 2\pi i
\left(\mathrm{Res}_{z=\rho_1} f(z)
+ \mathrm{Res}_{z=\rho_2} f(z)\right)$$
where $\rho_{1,2} = -1\pm 2i.$

The two residues are
$$\frac{\exp(1/2\log(\rho_{1,2}))}{2\rho_{1,2} + 2}.$$
Let $\rho_1 = \sqrt{5} \exp(i\theta)$ with $0\lt\theta\lt \pi$ so that
$\rho_2 = \sqrt{5} \exp(2\pi i - i\theta)$ with the chosen branch of the logarithm.
This gives
$$\mathrm{Res}_{z=\rho_1} f(z) =
\frac{5^{1/4} \exp(1/2 i\theta)}{2\rho_1+2}
= \frac{5^{1/4} \exp(1/2 i\theta)}{4i} $$
and
$$\mathrm{Res}_{z=\rho_2} f(z) =
- \frac{5^{1/4} \exp(-1/2 i\theta)}{2\rho_2+2}
= \frac{5^{1/4} \exp(-1/2 i\theta)}{4i}.$$
Collecting everything we obtain
$$\pi i \times
\frac{5^{1/4}}{2i} \cos(1/2\theta)
= \pi \frac{5^{1/4}}{2}  \cos(1/2\theta).$$

This is
$$\pi \frac{5^{1/4}}{2} \sqrt{\frac{1+\cos\theta}{2}}
= \pi \frac{5^{1/4}}{2} \sqrt{\frac{1-1/\sqrt{5}}{2}}
= \frac{\pi}{2} \sqrt{\frac{\sqrt{5}-1}{2}}.$$
A: Notice first that
$$
\int_0^\infty\frac{\sqrt{x}}{x^2+2x+5}\,dx\stackrel{x=u^2}{=}\int_0^\infty\frac{2u^2}{u^4+2u^2+5}\,du=\int_{-\infty}^\infty f(x)\,dx
$$
where
$$
f(z)=\frac{z^2}{z^4+2z^2+5}.
$$
The function $f$ has 4 simple poles at $a,-a,\bar{a}$, and $-\bar{a}$, with
$$
a=\sqrt{\frac{\sqrt{5}-1}{2}}+i\sqrt{\frac{\sqrt{5}+1}{2}}
$$
Given $R>|a|$, we denote by $\Delta(R)$ the bounded region of $\mathbb{C}$ whose boundary consists of the segment $[-R,R]$ and the upper half circle $C_R:=\{Re^{it}:\, 0\le t\le \pi\}$.
Since $a,-\bar{a}\in \Delta(R)$, thanks to the Residue Theorem we have:
\begin{eqnarray}
\int_{\Delta(R)}f(z)\,dz&=&2\pi i\left(\mathrm{Res}(f,a)+\mathrm{Res}(f,-\bar{a})\right)\\
&=&2\pi i\left[\frac{a^2}{2a(a-\bar{a})(a+\bar{a})}+\frac{\bar{a}^2}{2\bar{a}(a+\bar{a})(a-\bar{a})}\right]\\
&=&\pi i\left[\frac{a}{(a-\bar{a})(a+\bar{a})}+\frac{\bar{a}}{(a+\bar{a})(a-\bar{a})}\right]\\
&=&\frac{\pi i}{a-\bar{a}}=\frac{\pi i}{2i\Im a}=\frac{\pi }{\sqrt{2(\sqrt{5}+1)}}=\pi\frac{\sqrt{\sqrt{5}-1}}{2\sqrt{2}}.
\end{eqnarray}
It follows that
$$
\pi\frac{\sqrt{\sqrt{5}-1}}{2\sqrt{2}}=\int_{-R}^R f(x)\,dx+\int_0^\pi \frac{iR^3e^{i3t}}{R^4e^{4it}+2R^2e^{2it}+5}\,dt \quad \forall R>|a|.
$$
For every $R>\max\{|a|,\sqrt{+1\sqrt{6}}\}$ we have
$$
\left|\int_0^\pi \frac{iR^3e^{i3t}}{R^4e^{4it}+2R^2e^{2it}+5}\,dt \right|\le \int_0^\pi \frac{R^3}{R^4-2R^2-5}\,dt= \frac{\pi R^3}{R^4-2R^2-5} \to 0 \mbox{ as } R\to \infty,
$$
and we deduce that
\begin{eqnarray}
\int_0^\infty\frac{\sqrt{x}}{x^2+2x+5}\,dx&=&\int_{-\infty}^\infty f(x)\,dx=\lim_{R\to\infty}\int_{-R}^Rf(x)\,dx\\
&=&\lim_{R\to\infty}\left[\pi\frac{\sqrt{\sqrt{5}-1}}{2\sqrt{2}}-\int_0^\pi \frac{iR^3e^{i3t}}{R^4e^{4it}+2R^2e^{2it}+5}\,dt \right]\\
&=&\pi\frac{\sqrt{\sqrt{5}-1}}{2\sqrt{2}}.
\end{eqnarray}
