Why is dividing by $dx$ or some other differentitator in an integral considered taboo? Forgive my ignorance but I remember when I was learning calculus, I remember that when we integrate, we always multiply the differentiator to $F(x)$.
However, it was never explained to me why we always multiply the differentiator and not some other mathematical operation on the differentiator.
Bottom line, why can't we perform $\int{1/dx}$ or $\int{1\pm dx}$?
 A: For primitives, it is purely notation. For definite integrals, there is a reason. 
With the Riemann sums, you're multiplying. The integral is, roughly speaking, just the Riemann sum $$\sum_{i=1}^nf(t_i)(x_i -x_{i-1}) $$ as the lengths of each subinterval $[x_{i-1}, x_{i}]$ tend to $0$. In other words, they are all "infinitesimally" small. This is in quotes because in standard analysis, infinitesimals ($dx$, $dy$, etc) don't have a formal definition. 
What is the motivation for this definition? Well, say you have a car driving with a varying velocity and you want its displacement over a time interval. How is this done? You can approximate this to arbitrary precision by breaking the time interval up into small chunks and supposing the car travels at a constant velocity in these small chunks. Then it's just multiplication. As the chunks become smaller, the approximation is $< \epsilon$ the actual displacement. 
A: The intuitive understanding of the (Riemann) integral is that you are adding together the areas of a continuum of infinitesimally narrow rectangles. The $dx$ in this notation represents the width of those rectangles, while $f(x)$ is the height of the rectangle with base $[x,x+dx]$. Under this intuition, integrating a function without a $dx$ or with a $dx$ in the denominator would most reasonably result in an infinite value. 
A: $\int f(x) dx$ is just a symbol defining the integral of $f$. $dx$ tells you that the dummy variable of integration is $x$.
This is similar to asking why you cannot divide by $\sin$ to get
$$\sin x = y \implies x = \frac{y}{\sin} \text{ (woops!)}$$
That having been said, you might want to look at the Riemann–Stieltjes integral as it can handle the first case, if we interpret $1/dx \equiv d(1/x)$. However, I am fairly certain this is not what you want, nor is the notation $1/dx\equiv d(1/x)$ a very intuitive one.
