Integration by Parts confusion

Question

I am using this video to learn Laplace Transform. The example used is a fairly basic one:
$$ \int_{0}^{\infty}t.e^{-st}dt $$
Simple enough, you need to integrate by parts.
So, $u = t$ and $v^\prime = e^{-st}$

So, $u^\prime = 1$. Well, according to the video it is $dt$ and I can not understand why. If I differentiate $t$ w.r.t. $t$ then I get 1.

Is there something I am missing ?

Answer 1

If $u = t$, then we need $du$, not $\frac{du}{dt}$. $$u = t \implies \dfrac {du}{dt} = 1 \iff du = 1\cdot dt = dt$$

Just as with integration by substitution, when we designate what $u$ will be, we need to know what $du$ is:

$$u = t \implies du = 1\cdot dt = dt$$

Answer 2

One of the "tricky" notations in calculus within derivatives is ${dy \over dx}$. In many cases, this term is used as a literal fraction, such that the pieces $dy = y'$ and $dx = x'$ are separable and can be moved throughout an expression in the process of solving it. So when you write $u = t$ then take the derivative with respect to $t$, you are actually doing this: ${du \over dt} = {dt \over dt} = 1$. The video is taking the next step and saying that this implies $du = dt$.