Why does substitution work in integrals Let's say I have this integral:

$$\int_0^\infty e^{-t} \, dt$$

And I make the substitution:

$$t = nu$$

Then why I can say that:
$$dt = n\,du$$
and then put this into my integral like this:

$$\int_0^\infty e^{-nu}n\,du$$

What's happening in the background that allow this to be done?
I'm asking this because I don't feel confortable threating diferencial operators as fractions and I don't know why this can be done.
It's just a way of looking at the chain rule.

The chain rule is differentiation by substitution.

One can write $$\frac{d}{dx} f(g(x)) = f'(g(x)) g'(x),$$ or one can look at
$$
\frac{d}{dx} f(g(x))
$$
and then do this substitution:
$$
u = g(x),\qquad \frac{du}{dx} = g'(x).
$$
Then one writes
$$
\frac{d}{dx} f(g(x)) = \frac{d}{dx} f(u) = \frac{df(u)}{dx} = \frac{df(u)}{du}\cdot\frac{du}{dx} = f'(u)\cdot g'(x) = f'(g(x))\, g'(x).
$$

In the same way, when one sees
$$
\int f'(g(x)) g'(x) \,dx,
$$
one does the substitution
$$
u=g(x),\qquad \frac{du}{dx} = g'(x),\qquad du = g'(x)\,dx.
$$
Then one has
$$
\int f'(g(x)) g'(x) \,dx = \int f'(u)\,du = f(u)+C = f(g(x))+C.
$$

So integration by substitution is the chain rule in reverse, just as integration by parts is the product rule in reverse.
The substitution in integrals are the reverse of the chain rule in derivatives.

The following thread gives a fairly good explanation on why does the substitution works

Why does substitution work in antiderivatives?