$\int_0^\infty \frac{\sqrt{x}}{x^2+2x+5}\mathrm{d}x$ using Feynman's trick We are expected to solve an integral similar to $\int_0^\infty \frac{\sqrt{x}}{x^2+2x+5}\mathrm{d}x$ using contour integration, but I was wondering whether it would be possible to use the so-called Feynman's trick, i.e. differentiating under the integral sign. I tried using $$F(t) = \int_0^\infty \frac{\sqrt{x}e^{-t\sqrt{x}}}{x^2+2x+5}\mathrm{d}x$$
so that
$$F^\prime(t) = \int_0^\infty \frac{-xe^{-t\sqrt{x}}}{x^2+2x+5}\mathrm{d}x$$
Sadly, this does not work. So I'm looking for any starting point to approach this problem. I appreciate any help.

Well, you could solve this problem using Differentiation Under the Integral Sign, however I think that it wouldn't be an easy task and probably would end up in a tricky differential equation. Instead of it, I offer you a solution that just requires some substitutions.
$$I=\int_{0}^{\infty}{\frac{\sqrt x}{x^2+2x+5}dx}\overbrace{=}^{x\rightarrow\sqrt{5t}}5^{\frac{3}{4}}\int_{0}^{\infty}{\frac{\sqrt t}{{5\ t}^2+2\sqrt5t+5}dt}$$
Let's make some rearrangements:
$$I=\color{red}{\frac{2}{\sqrt[4]{5}}\int_0^{\infty}\frac{\frac{1}{2}\frac{1}{\sqrt t}}{\left(\sqrt t-\frac{1}{\sqrt t}\right)^2+\frac{10+2\sqrt5}{5}}dt}\overbrace{=}^{t\rightarrow \frac{1}{t}}\color{blue}{\frac{2}{\sqrt[4]{5}}\int_0^{\infty}\frac{\frac{1}{2}\frac{1}{t\sqrt t}}{\left(\sqrt t-\frac{1}{\sqrt t}\right)^2+\frac{10+2\sqrt5}{5}}dt}$$
Summing the integrals red and blue:
$$2I={\frac{2}{\sqrt[4]{5}}\int_0^{\infty}\frac{\frac{1}{2}\frac{1}{\sqrt t}+\frac{1}{2}\frac{1}{t\sqrt t}}{\left(\sqrt t-\frac{1}{\sqrt t}\right)^2+\frac{10+2\sqrt5}{5}}dt}\overbrace{=}^{\sqrt t-\frac{1}{\sqrt t}=u}\frac{2}{\sqrt[4]{5}}\int_{-\infty}^{\infty}\frac{du}{u^2+\frac{10+2\sqrt5}{5}}$$
$$2I=\frac{2}{\sqrt[4]{5}}\sqrt{\frac{5}{2\left(5+\sqrt5\right)}}\left[\arctan{\left(u\sqrt{\frac{5}{2\left(5+\sqrt5\right)}}\right)}\right]_{-\infty}^\infty$$
Hence:
$$I=\frac{\pi}{\sqrt[4]{5}}\sqrt{\frac{5}{2\left(5+\sqrt5\right)}}=\frac{\pi}{2\sqrt{\phi}}$$
