In my book (Young and Freedman's University Physics) they derive terminal velocity where and use these steps:
...we must find the relationship between velocity and time during the interval before the terminal speed is reached. We go back to Newton's second law, which we rewrite using $a_y = dv_y/dt$: $$m\frac{dv_y}{dt} = mg - kv_y$$ After rearranging terms and replacing $mg/k$ by $v_t$, we integrate both sides, noting that $v_y=0$ when $t=0$: $$\int_0^v \frac{dv_y}{v_y-v_t} = -{k\over m} \int_0^t dt$$ which integrates to $$\ln\frac{v_t-v_y}{v_t} = -{k\over m}t\mathrm{\qquad or \qquad}1-{v_y \over v_t}=e^{-(k/m)t}$$ and finally $$v_y=v_t[1-e^{-(k/m)t}]$$ Note that $v_y$ becomes equal to the terminal speed $v_t$ only in the limit that $t \to\inf$; the ball cannot attain a terminal speed in any finite length of time.
The step I don't understand is how they say that $$\int_0^v \frac{dv_y}{v_y-v_t} = \ln \frac{v_t-v_y}{v_t}$$ By my calculations, we have $$\int_0^v \frac{dv_y}{v_y-v_t} = \ln(v_y-v_t)\rvert_0^v = \ln(v-v_t)-\ln(-v_t) = \ln(\frac{v-v_t}{-v_t}) = \ln(\frac{v_t-v}{v_t}) \neq \ln\frac{v_t-v_y}{v_t}$$ How does the book arrive at its answer and how is my derivation wrong?
For that matter, what do $v$ and $t$ even represent, since their meanings were not defined?
