It is commonly taught that to integrate a function f(x), with respect to x, from x = a to x = b, one calculates:
$$\int_a^b f(x) \ dx$$
We add dx at the end of the integral to show that we are summing up the area of an infinite number of thin blocks, where the height of each block is given by f(x) and the width is given by dx. In this way, we can calculate the area bounding by the x axis, f(x), x = a, and x = b.
However, what about integrals that were meant to calculate certain other quantities, such as arc length? The formula for the arc length of a function ends with dx, but dx exists for another reason. This reason is that when deriving the arc length formula, $$(\Delta\ x)^2$$ appears under a square root, leaving $$\Delta\ x$$
Then, as $$\Delta\ x$$ becomes the infinitesimal dx, and the summation becomes an integral, we see the formula become an integral that ends with dx. This is not the same as an integral that ends with dx in order to find the area under a curve. Yet somehow, all integrals, whatever their purposes, end with dx. What is the general meaning of dx in an integral, seeing that all integrals must end in dx?
So, if you have a function $f$, the arc length formula tells you that the length of $(x,f(x))$ for $x$ going from $a$ to $b$ is $$ \int_a^b \sqrt{1+f'(x)^2}\,\mathrm{d}x $$ And actually, it is also the area under the function $\sqrt{1+f'(x)^2}$, this is the remarkable fact that the theorem tells you. Then, not all integrals end by $\mathrm{d}x$, which correspond to integrals with Lebesgue measure. More generally you can have $$ \int f(x)\,\mu(\mathrm{d}x), $$ and in the case of the Lebesgue measure, you have the common notation $∫ f$ instead of $∫ f(x) \mathrm{d}x$.
Moreover, in the case when $g$ is a function of bounded variations, then $μ = g'$ is a measure and there is the Stieltjes integral notation $$ ∫f(x)\,\mu(\mathrm{d}x) = ∫f(x)\,\mathrm{d}g(x) = ∫f\,\mathrm{d}g $$