Is $\int \frac{{dx}^2}{dy}$ valid? It appears to me that
$\int \frac{{dx}^2}{dy}$ can be rewritten as
$\int \frac{dx}{dy}\cdot{}dx$ which in turn can be rewritten as
$\int f^{-1}{^\prime}(y)\cdot{}dx$. (it is assumed that $y=f(x)$ although y may not be a function of x; multiple solutions for y may exist for a given value of x if y is, say, $\pm\sqrt{x}$).
Last time I checked, $\frac{dy}{dx}$ is a psuedo fraction, and taking the integral of the derivative of the inverse function of y as a function of x with argument y is possible. But I'm new to integral calculus so I could be mistaken :)
So for a more tangible example, assume $y=x^2$
then $\frac{dx}{dy}=\frac{1}{\frac{dy}{dx}}=\frac{1}{2x}=\frac{1}{2\sqrt{y}}=\frac{d}{dy}\left(\sqrt{y}\right)=\frac{d}{dy}\left(x\right)$
so $\int \frac{dx}{dy}\cdot{}dx = \int\frac{1}{2x}\cdot{}dx=\frac{ln(x)}{2}+C$
What my question is, is can you just do that? rewrite $dx\cdot{}dx$ as ${dx}^2$? I'm fairly confident the rest of my explanation is logically sound. If you feel you know the answer to my original question, feel free to answer it! If you think I am mistaken somewhere, please explain precisely where and don't just accuse me of nonsensically tinkering with mathematical syntax that "doesn't work that way" without explaining where exactly I'm mistaken.

Your interpretation of $dx^2$ as $(dx)(dx)$ may get confused with $ d(x^2)$. 
Otherwise your thought process is straightforward and worths further investigation.
You example makes perfect sense. 
