I am currently learning how to find antiderivatives using the "$u$-substitution" or "integration by substitution" method. A key component of this is setting some expression in the indefinite integral as "$u$", and then also finding $du/dx$.
How can we then write $du = dx \cdot $ (some expression)? Isn't $du/dx$ defined as the derivative of $u$ and is not a fraction?
On
From my point of view it is better to think of substituting $x$ by $x(u)$. The substitution, however, changes the shape of the function and therefore the integral. The picture below illustrates this fact. There is graphical approach to show why you need a special replacement of $dx$: You can think of compensating for the density by multiplying by it, namely $dx/du$. However, you may find it helpful to read the full explanation on Insight Things.
From a simplistic viewpoint, you can view this as a convenient mnemonic: a notation that helps you remember what's true. Once you've learned that, given dependent variables $u$ and $x$ that satisfy $u = g(x)$, you can choose the antiderivatives so that
$$ \int f(u) \, \mathrm{d}u = \int f(g(x)) g'(x) \, \mathrm{d}x $$
then it is much easier to remember the rule via implicit differentiation of differentials: that $\mathrm{d}u = \frac{\mathrm{d}u}{\mathrm{d}x} \, \mathrm{d}x$.
From a deeper viewpoint, once you recognize the patterns going on here, you'll want to form a concept of a differential $\mathrm{d}x$ as a new kind of mathematical object that really exists, and is something you can manipulate.
One can then define an integral not to be the integral of a function, or as the integral of an expression with respect to one of its variables, but instead you take integrals of differentials.
From this more sophisticated viewpoint, that's what integrals are supposed to be, and that the integral you learned in introductory calculus was just a simplification that is easier to teach and requires less sophistication to discuss rigorously.