What is $\int{\sqrt{dx}}$ How would I compute the integral $\int{\sqrt{dx}}$?
I would suppose it is recursive like this:
$\int{(dx)^2} = (x + C)dx$
Is this a well formed mathematical term? I can imagine it is because
$\iint{(dx)^2} = \frac{x^2}{2} + Cx + D$ and thus $\frac{d\iint{dx^2}}{dx} = x + C$ and then $\int{dx^2} = (x + C)dx$.
 A: $\int{\sqrt{dx}}$ is an ambiguous symbolism, as already pointed out by several posters.
The tone of the question draw me to think that you are looking for something like "fractionnal calculus", i.e. antiderivative and derivative of non-integer degree. In this sense, it should be better to raise the question on this form :
How would I compute $\frac{d^\nu}{dx^\nu}f(x)$ ? where $\nu=-\frac{1}{2}$ and $f(x)=1$ in your example. 
For information, see:
http://mathworld.wolfram.com/FractionalDerivative.html
http://mathworld.wolfram.com/FractionalIntegral.html
Your particular case corresponds to the definition of "semi-integral", i.e. fractionnal integral of order 1/2 , as shown here :
http://mathworld.wolfram.com/Semi-Integral.html
The result is :
$$\frac{d^{-1/2}}{dx^{-1/2}}(1)=2\sqrt{\frac{x}{\pi}}$$
A paper for general public is available here :
https://fr.scribd.com/doc/14686539/The-Fractional-Derivation-La-derivation-fractionnaire
Note that your writting :
$\iint{(dx)^2} = \frac{x^2}{2} + Cx + D$ and thus $\frac{d\iint{dx^2}}{dx} = x + C$ and then $\int{dx^2} = (x + C)dx$ 
is not correct for two reasons : 
First, $\frac{d\iint{dx^2}}{dx}$ is not equal to $\int{dx^2}$ but is equal to $\int{dx}$.
Second : The right term cannot be an infinitesimal such as $(x + C)dx$. The $dx$ is too much.
