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.