Evaluating $\int_0^{\infty} \frac{\sqrt{x}}{x^2+2x+5} dx$ using complex analysis how do I compute
$$\int_0^{\infty} \frac{\sqrt{x}}{x^2+2x+5} dx$$
with complex analysis?
I feel like im calculating the residue wrong and I cant get to the answer correctly.
I tried to branch cut the real $0 \rightarrow \infty$ but I feel like im doing it wrong. any help is appriciated.
additional information: 
thank you for the input everyone it is very helpful.
i did come down to calculating the integral 
$$\int_0^{\infty} \frac{\sqrt{x}}{x^2+2x+5} dx = i\pi [Res(f,z_1=-1+2i)+Res(f,z_2=-1-2i)]$$
Then given answer to this question is $\frac{\pi}{2}\sqrt{\frac{\sqrt{5}-1}{2}}$
I was just simply calculating 
$i\pi [Res(f,z=-1+2i)+Res(f,z=-1-2i)] = i\pi \left(\frac{\sqrt{z_1}}{2z_1+2}+\frac{\sqrt{z_2}}{2z_2+2}\right)$
Solving for
$\frac{\pi}{4} \left(\sqrt{-1+2i}-\sqrt{-1-2i}\right)$
I get $\frac{\pi}{2}\sqrt{\frac{-\sqrt{5}-1}{2}}$ and I still dont know what I am doing wrong for that one sign error.
 A: The integral along the contour just above the real axis would be
$$
\int_0^R\frac{\sqrt{x}}{x^2+2x+5}\mathrm{d}x\tag{1}
$$
The integral along the contour circling the complex plane at a radius of $R$ would be bound by
$$
\begin{align}
\int_0^{2\pi}\frac{\sqrt{R}}{R^2-2R-5}R\,\mathrm{d}\theta
&\stackrel{\hphantom{R\to\infty}}{\le}\frac{2\pi}{R^{1/2}-2R^{-1/2}-5R^{-3/2}}\\
&\stackrel{R\to\infty}{\to}0\tag{2}
\end{align}
$$
The integral along the contour just below the real axis would be
$$
\int_R^0\frac{-\sqrt{x}}{x^2+2x+5}\mathrm{d}x\tag{3}
$$
Adding up the pieces $(1)$, $(2)$, and $(3)$, and letting $R\to\infty$, we get
$$
\begin{align}
2\int_0^\infty\frac{\sqrt{x}}{x^2+2x+5}\mathrm{d}x
&=2\pi i\left[\vphantom{\frac{\sqrt{-1-2i}}{-4i}}\right.\underbrace{\frac{\sqrt{-1-2i}}{-4i}}_{\begin{array}{}\text{Residue at}\\z=-1-2i\end{array}}+\underbrace{\frac{\sqrt{-1+2i}}{4i}}_{\begin{array}{}\text{Residue at}\\z=-1+2i\end{array}}\left.\vphantom{\frac{\sqrt{-1-2i}}{-4i}}\right]\\
&=2\pi i\left[\frac{-\sqrt{\phi-1}+i\sqrt\phi}{-4i}+\frac{\sqrt{\phi-1}+i\sqrt\phi}{4i}\right]\\[12pt]
&=\frac\pi{\sqrt\phi}\tag{4}
\end{align}
$$
and therefore,
$$
\int_0^\infty\frac{\sqrt{x}}{x^2+2x+5}\mathrm{d}x=\frac\pi{2\sqrt\phi}\tag{5}
$$
