Another approach to evaluating the integral with appealing to either Cauchy's Integral Formula or the Residue Theorem relies on Cauchy's Theorem.
If $f(z)$ analytic within a open region (simply connected) in the complex plane, then for any contour (with finite length) $C$ contained in that open region,
$$\oint_C f(z) dz=0$$
Now, in the problem at hand, $f(z)$ is not analytic in the region encircled by $C$. However, we may evaluate the integral on a "deformed" contour $C'$ that does not enclose any singularity. Then,
$$\oint_{C'} f(z)dz=0$$
If we deform $C$ by adding a "key-hole" contour that "cuts out" the singularity with a small circle $\gamma$ of radius $\epsilon$, centered at the singular point then
$$\oint_{C'} f(z)dz=\oint_C f(z)dz-\oint_{\gamma}f(z)dz=0$$
where the opposing contributions from the "key length" integrations annihilate one another.
Thus, this reduced the problem to evaluating the integral over $\gamma$. For this problem,
$$\begin{align}
\oint_{\gamma}f(z) dz&=\lim_{\epsilon \to 0}\left(\int_0^{2\pi}\frac{i\epsilon e^{i\phi}d\phi}{(i+\epsilon e^{i\phi})^2+1}\right)\\\\
&=\lim_{\epsilon \to 0}\left(\int_0^{2\pi}\frac{i\epsilon e^{i\phi}d\phi}{2i\epsilon e^{i\phi}+\epsilon^2 e^{i2\phi}}\right)\\\\
&=\pi
\end{align}$$
Yet another way: we may remove the branch point at the origin by setting $x=z^4$, leading to
$$ I = \int_{0}^{+\infty}\frac{dx}{\sqrt[4]{x}(1+x^2)}=\int_{0}^{+\infty}\frac{4z^2\,dz}{1+z^8}=\int_{0}^{1}\frac{4z^2\,dz}{1+z^8}+\int_{0}^{1}\frac{4z^4}{z^8+1}\,dz$$
then to
$$ I = 4\int_{0}^{1}\frac{(z^2+z^4)(1-z^8)}{1-z^{16}}\,dz=4\sum_{n\geq 0}\left[\tfrac{1}{16n+3}+\tfrac{1}{16n+5}-\tfrac{1}{16n+11}-\tfrac{1}{16n+13}\right].$$
Now the Dirichlet $L$-series appearing in the RHS can be computed by recalling that
$$ \sum_{n\geq 0}\left[\frac{1}{an+b}-\frac{1}{an+(a-b)}\right]=\frac{\pi}{a}\cot\left(\frac{\pi b}{a}\right) $$
holds by the reflection formula for the $\psi$ function / Herglotz trick. We get
$$ I = \frac{\pi}{4}\left[\cot\left(\frac{3\pi}{16}\right)+\cot\left(\frac{5\pi}{16}\right)\right]=\color{red}{\frac{\pi}{\sqrt{2+\sqrt{2}}}}. $$
Best Answer
Since the wikipedia page on methods of contour integration is flawed (check the talk section), consequently, most explanations concerning the calculus of integrals using a "keyhole contour" on the web are flawed or not rigorous enough, in my opinion.
I'll try to explain how to do this, or at least the part that is not well explained.
As mentioned above, you can avoid all these difficulties by substituting $x=u^2$ to get rid of the square root.
Let $r$,$R$ and $\alpha$ be three positive numbers, with $0<r<R$, and $0<\alpha<\pi$.
Let $U=\Bbb C\setminus \Bbb R_+$ the branch cut of $\Bbb C$ along the positive real axis.
Since $U$ is star-shaped with respect to any real negative number, it's a simply connected domain of $\Bbb C$.
Let $\gamma=\gamma_1+\gamma_2+\gamma_3+\gamma_4$ be the contour showned on the picture below (the red part is the branch cut) :
$\hspace{4cm}$
More precisely, we have :
$\hspace{5cm}\begin{array}{ccccl}\gamma_1&:&[r,R]&\longrightarrow& U\\ &&t&\longmapsto &te^{i\alpha}\end{array}$
$\hspace{5cm}\begin{array}{ccccl}\gamma_2&:&[\alpha,2\pi-\alpha]&\longrightarrow& U\\ &&t&\longmapsto &Re^{it}\end{array}$
$\hspace{5cm}\begin{array}{ccccl}\gamma_3&:&[-R,-r]&\longrightarrow& U\\ &&t&\longmapsto &-te^{-i\alpha}\end{array}$
$\hspace{5cm}\begin{array}{ccccl}\gamma_4&:&[\alpha,2\pi-\alpha]&\longrightarrow& U\\ &&t&\longmapsto &re^{-it}\end{array}$
We have $\mathrm{Supp}(\gamma)\subset U$, and we can define a holomorphic square root on $U$ like this :
$\hspace{1cm}$ If $z=re^{i\theta}$ with $r>0$ and $0<\theta<2\pi$, then $\sqrt{z}=\sqrt{r}e^{i\frac{\theta}2}$.
You can notice that the image of this function is the half open plane $\{z\in\Bbb C :\mathrm{Im}(z)>0\}$.
Now let : $\hspace{1cm} f:z\longmapsto \dfrac{\sqrt{z}}{z^2+2z+5}=\dfrac{\sqrt{z}}{g(z)}$
Since the function $g:z\longmapsto z^2+2z+5$ is a polynomial, it's holomorphic on $U$, and since it's non constant, the function $f$ is meromorphic on $U$ as a quotient of two holomorphic functions with a non constant zero denominator.
The zeros of $g$ are $z_\varepsilon=-1+2\varepsilon i$ with $\varepsilon=-1$ or $1$, and they are the poles of $f$. Since $\mid z_\varepsilon\mid=\sqrt 5$ and $\mathrm{Re}(z_\varepsilon)<0$, we will assume that $0<r<\sqrt 5<R$ and $0<\alpha<\dfrac{\pi}2$ from now on.
This way, the two poles of $f$ lie in the interior of $\gamma$, and the winding number around them with respect to $\gamma$ is 1.
By the residue theorem, we get :
$$\int_\gamma f(z)dz=2i\pi \sum_{\varepsilon\in\{-1;1\}} \mathrm{Res}(f,z_\varepsilon)$$
Let's calculate the residues of $f$ at $z_\varepsilon$, for $\varepsilon=-1$ and $1$.
They are simple poles of $f$, so : $$\mathrm{Res}(f,z_\varepsilon)=\lim_{z\to z_\varepsilon} (z-z_\varepsilon)\dfrac{\sqrt z}{g(z)}=\lim_{z\to z_\varepsilon} \left[\dfrac{z-z_{\varepsilon}}{g(z)-g(z_\varepsilon)}\times\sqrt{z}\right]=\dfrac{\sqrt{z_\varepsilon}}{g'(z_\varepsilon)}$$
We have $g'(z_\varepsilon)=2z_\varepsilon+2=2(z_\varepsilon+1)=4\varepsilon i$.
Letting $z_\varepsilon=(a+ib)^2$ with $(a,b)\in\Bbb R^2$ quickly leads to $\displaystyle\sqrt{z_\varepsilon}=\varepsilon\sqrt{\frac{\sqrt 5-1}2}+i\sqrt{\frac{\sqrt5+1}2}$ (remember that $\mathrm{Im}(\sqrt{z_\varepsilon})>0$)
So we get :
$$2i\pi\sum_{\varepsilon\in\{-1;1\}}\mathrm{Res}(f,z_\varepsilon)=\dfrac{2i\pi}{4i}(\sqrt{z_1}-\sqrt{z_{-1}})=\pi\sqrt{\frac{\sqrt 5 -1}2}$$
We found that : $$\int_\gamma f(z)dz=\sum_{k=1}^4 \int_{\gamma_k} f(z)dz=\pi\sqrt{\frac{\sqrt5-1}2}$$
No we will prove that these four integrals have a limit when $r$ and $R$ are fixed and $\alpha\longrightarrow 0$.
Of course, you can't just put $\alpha=0$ like this, because $f$ isn't defined on $\Bbb R_+$ !
The function $\displaystyle\left[0,\dfrac\pi 2\right]\times\left[r,R\right]\ni (\alpha,t)\mapsto \dfrac{\sqrt te^{i\frac{3\alpha}2}}{g(te^{i\alpha})}$ is continuous on a compact set so we can let $\alpha\rightarrow 0$ in the integrand (by dominated convergence) to get the limit $\displaystyle\int_r^R \dfrac{\sqrt{t}}{t^2+2t+5}dt$.
In the same manner, you can show that $\displaystyle\int_{\gamma_3} f(z)dz$ tends to the same limit.
Since $\displaystyle\left|1_{[\alpha,2\pi-\alpha]}(t) \frac{iR^{\frac32}e^{i\frac{3t}2}}{g(Re^{it})}\right|\leq \frac{R^{\frac{3}{2}}}{\mid g(Re^{it})\mid}\leq\frac1{\sqrt{R}}$ for all $(\alpha,t)\in\left[0,\dfrac{\pi}2\right]\times\left[0,2\pi\right]$, we find that $\displaystyle\int_{\gamma_2}f(z)dz$ tends to $\displaystyle\int_0^{2\pi}\frac{iR^{\frac 32}e^{i\frac{3t}2}}{g(Re^{it})}dt$ (again using dominated convergence) when $\alpha$ tends to $0$.
In the same manner, you can prove that $\displaystyle\int_{\gamma_4} f(z)dz$ tends to $\displaystyle\int_0^{2\pi}\frac{ir^{\frac32}e^{-i\frac{3t}2}}{g(re^{-it})}dt$
So we just proved that :
$$2\int_r^R \frac{\sqrt t}{t^2+2t+5}dt+\int_0^{2\pi}\frac{iR^{\frac 32}e^{i\frac{3t}2}}{g(Re^{it})}dt+\int_0^{2\pi}\frac{ir^{\frac32}e^{-i\frac{3t}2}}{g(re^{-it})}dt=\pi\sqrt{\frac{\sqrt 5 -1}2}$$
whenever $0<r<\sqrt 5<R$, so you can write, in such cases :
$$\int_r^R \frac{\sqrt t}{t^2+2t+5}dt=\frac{\pi}2\sqrt{\frac{\sqrt 5-1}2}-\frac 12\left(\int_0^{2\pi}\frac{iR^{\frac 32}e^{i\frac{3t}2}}{g(Re^{it})}dt+\int_0^{2\pi}\frac{ir^{\frac32}e^{-i\frac{3t}2}}{g(re^{-it})}dt\right)$$
Like the others before, I leave the rest to you : you have to prove that the limits of the two integrals in the parentheses are $0$ when $r$ tends to $0$ and $R$ tends to $+\infty$