calculus - A really basic integration question concerning differentials

I'm really, really confused with this. Please, please help me.
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My Calculus teacher taught me that the integral symbol and the differential with respect to which we are integrating are like parenthesis. He told us to think of the integral sign as an “open parenthesis” and the $dt$ as a “close parenthesis”.
If we were to integrate any function of $t$, say $v(t)$, we have to put the integral sign on the left of $v(t)$, and the differential $dt$ on the right of $v(t)$ ie $\int v(t) dt$ $$$$
In physics, while while deriving equations of motion, our physics teacher did this: since v is a linear function of t, $$dv(t)/dt=a(constant)$$ $$\Rightarrow dv= a dt$$
She then simply put an integral sign on LHS and RHS to integrate and then got $$\int dv= \int a dt$$

$$v=at+C$$
But this does not fit into what my Calculus teacher had taught us. As per what he has taught:
$$ dv= a dt$$
We now have to add an integral sign and another differential on either side of the expressions in LHS and RHS respectively. Only then can we integrate.$$$$
Could somebody please explain this idea of the integral sign as an “open parenthesis” and the $dt$ as a “close parenthesis”? Please could you explain how this is applied to the physics example?

Answer

No doubt you already know the fact $$\int dy=y+c$$
where $y$ can be any (sensible) thing whatsoever. You probably also know that $$\int \frac{dy}{dt} dt=y+c$$
since integration undoes differentiation (almost), by the fundamental theorem of calculus (you might not recognise that name, even if you recognise the fact). Indeed, you probably also know the chain rule which states $$\frac{dy}{dt}=\frac{dy}{du}\frac{du}{dt}$$ where $y$ and $u$ are any sensible quantities. People who aren't pure mathematicians (and especially physicists) tend be abit risqué and like to treat differentials like they're numbers, and write things like $$dv=\frac{dv}{du}du$$
since we can "divide by $dt$" (or by any other differential) and get the chain rule. While what your teacher is doing is absolutely incorrect, it's safe because of these theorems working in the background, and lets your notation be a bit more elegant.

In the example you give, what is really happening is: $$\frac{dv}{dt}=a \implies \int \frac{dv}{dt}dt=\int a dt \implies v=at+c$$

Which your teacher has abbreviated to $$dv=adt \implies \int dv = \int adt \implies v=at+c$$

P.S. At its core this isn't a basic question, and you shouldn't be worried about being confused by it.