The example given is a physics application but that's not the point. I don't understand what the question has to do with this case of integration. I do know that that's a main use of integration but here he gave $t$ as an integral of $v$. Shouldn't I differentiate both sides relative to $v$ to get $\frac{dt}{dv}$ then use the inverse rule to get $\frac{dv}{dt}$, then integrate relative to $t$?
That's my line of thinking; either have it corrected or his answer explained.
$t = \int \frac 1{20-v} dv$ so you want to differentiate both sides with respect to v.
$\frac {dt}{dv} = \frac {1}{20-v}$
and invert it.
$\frac {dv}{dt} = 20-v$
Giving you a linear diff eq.
$v = C_1 e^{-t} + 20$
$v(0) = 0$
$v = (-20) e^{-t} + 20$
that will work, just as well.