This is a very problematic notation, I'll first talk informally and then give you the right way to do things. The point is that it has to do with change of variables. For example, if you have $y=(ax+b)^2$ then if we set $u = ax+b$ we would have $y=u^2$. Thus, if we change in $x$, there is a change ocurring in $u$ and this will give you a change in $y$ that you can calculate with the chain rule.
In the usual notation for derivatives, the chain rule would be stated as:
$$\frac{dy}{dx}=\frac{dy}{du} \frac{du}{dx}.$$
Now, this notation is confusing, this notation is bad, and although everyone should know it to understand when read books and articles using it, people should really move to the modern notation. I'll explain why: it all has to do with the notion of composition of functions.
If we have two functions $f : \mathbb{R} \to \mathbb{R}$ and $g : \mathbb{R} \to \mathbb{R}$ one can form the composition $g \circ f : \mathbb{R} \to \mathbb{R}$ that is defined by $(g\circ f)(x)=g(f(x))$ so, the composition is the result of applying $g$ to the result of applying $f$ to $x$.
In this notation, the chain rule is written as $(g\circ f)'(x)=g'(f(x))f'(x)$ and this notation is much better because it doesn't carry any ambiguities. It says "to take derivative of a composition, take the derivative of $g$ and of $f$ normally, then apply the derivative of $g$ at $f(x)$ and multiply by the derivative of $f$ at $x$".
The example I gave you would have $g(x)=x^2$, $f(x)=ax+b$ and so
$$(g\circ f)(x)=g(f(x))=(ax+b)^2$$
To differentiate, we have $g'(x) = 2x$ and $f'(x)=a$ so $g'(f(x))=2(ax+b)$ and then:
$$(g\circ f)'(x)=2a(ax+b)$$
The usual notation carries ambiguities. First, notice that the function being defined doesn't depend on the letter used: the letter is just a symbol! So, writing $f(x)=x^2$ or $f(u)=u^2$ is the exact same thing, $x$ and $u$ are just placeholders for real numbers.
In the usual notation, the left hand side talks about the derivative of $y=(ax+b)^2$ and the right hand side about the derivative of $y=u^2$. So, $y$ is representing two different functions and this confuses a lot of people.
All of this is really confusing, and the rigorous framework exists because we don't want ambiguities: learning something rigorous can seem a little harder, but you will be able to understand without ambiguities like that. In that case, my suggestion is this : get the book Calculus by Michael Spivak, it will teach you how to think about Calculus in a logical way, running away from this kind of confusion.
I hope this helps you somehow. Good luck!
EDIT: The notation $f: \mathbb{R} \to \mathbb{R}$ just means that $f$ is a function with domain $\mathbb{R}$ and codomain $\mathbb{R}$, in other words, $f$ maps real numbers into real numbers. In general, given sets $A$ and $B$ one function that takes elements of $A$ into elements of $B$ is written $f : A \to B$.
Definite integration of continuous, single-variable functions gives you the area between the function and the $x$-axis. This area is a value, not a function.
Indefinite integration results in a function that describes the area in terms of the independent variable $x$. The result, $F(x)$ of an indefinite integral of a primitive function $f$ gives you a function which, given a value $X$, returns the area under $f$ between $0$ and $X$.
Integration doesn't give you an "equation" of higher "dimension". Indefinite integration gives you a function that can be used to compute area; definite integration gives you a single number that describes area. Now, integration of polynomials does in fact result in a polynomial of a higher degree, which has a natural link to dimensionality, but the two should not be conflated.
When you extend to functions of higher order, it becomes more complicated. This is because the notion of the differential becomes a bit more complicated for functions of more than one variable. For single-variable functions, the differential $dx$ is pretty easy to deal with; for functions of two variables, we need to formally describe what we mean by these differentials, such as $dA$, a differential "area" element, or the product $dx\ dy$. There are different, but equivalent, notions of these that you will encounter if you ever study real analysis in detail. I won't go into them now.
It may also surprise you to know that there are many different kinds of integrals. The integral operation is something very deep and complicated. However, they can all be tied back into the same notion of "infinitely summing" over something. It's just that there are many pathways to describe more or less the same thing!
Best Answer
The most transparent way of understanding $dx$ and $du$ in integration is as infinitesimal increments, and the integral - in terms of an infinite sum of infinitesimals. When you introduce a change of variables, for example $u=10x$, the infinitesimal increments transform accordingly, in this case $du=10dx$. Of course, the bounds of integration change accordingly: if $x$ varies from $a$ to $b$, the new variable $u$ will vary from $10a$ to $10b$. Of course, similar rules apply for nonlinear changes of variable.