Evaluate $\int_0^{\infty}\frac{\sqrt{x}}{x^2+1}dx$ without Complex Analysis I found  this question on a Complex Analysis Qualifying Exam:

How would you evaluate: $$\int_0^{\infty}\frac{\sqrt{x}}{x^2+1}dx$$

I am interested in all methods including Complex Analysis, I'm just less familiar with it. 

For instance, since the integral is from $0$ to $\infty$ do we only consider the $x-i$ factor in $x^2+1=(x-i)(x+i)$ since $i$ would be the only root in the upper half plane?
Write $x=\tan t$ so your integral is $$\int_0^{\pi/2}\tan^{1/2}tdt=\frac12\operatorname{B}\left(\frac14,\,\frac34\right)=\frac{\pi}{2}\csc\frac{\pi}{4}=\frac{\pi}{\sqrt{2}}.$$See here and here if any of the theory I used is unfamiliar.
Let $x^2=y$ we obtaine : 

$\int_0^{\infty}\frac{x^{-\frac{1}{4}}}{2(1+x)}dx$ 

Then we use beat function : 

$\beta(\frac{3}{4},\frac{1}{4})$