Writing du in terms of dx (Integrating using U substitution) suppose we have the following:
$y=(2x+1)^7$
and we want to integrate it, then we set $u = 2x+1$
but then.....$du = 2dx$..... logically speaking now, du and dx are only "pointers" that show you which variable you should integrate the expression with respect to. How can these be divided and multiplied?
 A: You're right in thinking of the $du$ and $dx$ in derivatives like $\frac{du}{dx}$ and integrals $\int_\alpha^\beta f(x) \,dx$ as purely notational. Without going pretty far out of our way to define them, they are just notational indicators of the functions being differentiated and variables with respect to which we differentiate and integrate.
So...why can we sometimes treat them as variables and $\frac{du}{dx}$ as a fraction? The answer is ultimately that the notation we use for derivatives and integrals is designed in such a way that some common manipulations like the $u$-substitution "just work out" the way we'd like them to. (Note: I mean to suggest that you can think of them as notation for now, but there are ways to make calculations like the one you mentioned rigorous. See the "Caveat" at the end of my answer for an example.)
Let's return to your example. In its basic form, substitution in an integral looks something like this:
$$
    \int_a^b f(u(x))\,u'(x)\,dx = \int_{u(a)}^{u(b)} f(u)\,du.
$$
(This assumes certain properties of $f$, $u$, and $u'$, but I'll avoid them until later.) In practice, you apply this formula by


*

*Beginning with some integral $\int_a^b g(x)\,dx$

*Realizing that you can write $g(x) = f(u(x))\,u'(x)$ for suitable functions $f$ and $u$

*Performing the substitution.


It sounds like you've been told to do this by following the following steps:


*

*Identify an intermediate function $u(x)$ that you can substitute.

*Compute $\frac{du}{dx} = \text{(some expression in $x$)}$

*"Multiply by $dx$" to get $du = \text{(some expression in $x$)}\,dx$

*If it's a definite integral (that is, if there are bounds), compute the value of $u$ at the lower and upper $x$-bounds. These are the bounds of your new integral.


The only step here that's an abuse of notation is when we multiply by $dx$ in 3. Looking back at the substitution formula I wrote above, you're finding a way to write part of the integrand in the form $f(u(x))$ and the rest of the integrand and the differential $dx$ as
$$
    u'(x) \,dx = \text{(some expression in $x$)}\,dx.
$$
If you'd like, you can go about all your substitutions like I was taught and not ever use step 3., but it's a convenient "accounting" trick to know, especially when you start looking at harder substitution problems and, later, things like integration by parts.

Bonus: Deriving the substitution formula
In case you're interested, you can actually prove that the substitution formula is true using what you've (hopefully!) already seen in class:
Theorem: (Integration by Substitution)
If $u$ is a differentiable function defined on the interval $[a,b]$, with a derivative $u'$ that is integrable on $[a,b]$, and if $f$ is a continuous function on an interval containing the range of $u$, then
$$
    \int_a^b f(u(x))\,u'(x)\,dx = \int_{u(a)}^{u(b)}f(u)\,du.
$$
You can think of the proof of this as just "reversing" the Chain Rule used when taking derivatives. The key points of the proof are as follows:


*
    
*
        The assumptions made in the theorem ensure that the integrals
        $$
        \int_\alpha^\beta f(u(x))\,u'(x)\,dx
        \qquad\text{and}\qquad
        \int_{u(\alpha)}^{u(\beta)}f(u)\,du
    $$
        are defined for all $\alpha$ and $\beta$ in the interval $[a,b]$. Accordingly, we just need to prove that they're equal when $\alpha = a$ and $\beta = b$.
    
    
*
        Since $f$ is continuous, it has an antiderivative $F$ on an interval containing $\operatorname{Range}(u)$. So by definition $F'(u) = f(u)$ for all $u$ in $\operatorname{Domain}(f)$, and one half of the Fundamental Theorem of Calculus tells us that
\[
    \int_{u(a)}^{u(b)} f(u)\,du = F(u(b)) - F(u(a)).
\]
    
    
*
        Note that the function $F(u(x))$ includes functions that are closely related to those in the other integrand $f(u(x))\,u'(x)$. The Chain Rule shows how these functions are related:
        $$
        [F(u(x))]' = F'(u(x))\,u'(x) = f(u(x))\,u'(x).
    $$
In other words, the function $F(u(x))$ is an antiderivative of $f(u(x))\,u'(x)$! So a second application of the Fundamental Theorem of Calculus shows that
        $$
        \int_a^b f(u(x))\,u'(x) = F(u(x))\,\bigg|_{x = a}^{x = b} = F(u(b)) - F(u(a)).
    $$
    
    
*Compare the last equations in numbers 2. and 3. to complete the proof.


Caveat
As observed in the comments, to say that differentials are purely notational isn't really true for long. You can define them more precisely and in a way that makes these formal calculations rigorous. One example of this (that you probably don't need to worry about for now) is found in the definition of the Riemann-Stieltjes integral.
