## time it takes for submerged object to rise to surface

$$\frac {1} {a} \int_0^v \frac {dv} {1 - \alpha^2 v^2} = \frac {1} {2a} \int_0^v \left[ \frac {1} {1 - \alpha v} + \frac {1} {1 + \alpha v}\right]dv = \frac {1} {2\alpha a} \left[ -ln (1 - \alpha v) + ln(1 + \alpha v)\right]_0^v = \frac {1} {2\alpha a} \ln \frac {1 + \alpha v} {1 - \alpha v}$$ Using Euler's formula for hyperbolic functions, we can see immediately that the latter is $\frac {1} {\alpha a} \tanh^{-1} \alpha v$, so $$\frac {1} {\alpha a} \tanh^{-1} \alpha v = \int_0^t dt = t$$

 The second integration is even easier: $$\frac {1} {\alpha} \int_0^t \tanh \alpha a t dt =\frac {1} {\alpha} \int_0^t \frac {\sinh \alpha a t} { \cosh \alpha a t }dt = \frac {1} {\alpha} \int_0^t \frac { (\frac {\cosh \alpha a t} {\alpha a})'} { \cosh \alpha a t }dt =$$$$= \frac {1} {\alpha^2 a} \left[\ln \cosh \alpha a t\right]_0^t = \frac {1} {\alpha^2 a} \ln \cosh \alpha a t = \int_{x_0}^{x} dx = x - x_0$$
 Cool, thanks for sharing the way you worked them out! I think I prefer the u substitution and tanh^-1 identity over using logarithms and Euler's formula, but mostly just because I already have all the hyperbolic identities and their inverses memorized from differentiation. That's still an interesting way to see it done though, and I'm going to try to remember how to do it. I saw that last identity yesterday (Euler's formula) in my book when looking for hyperbolic identities, but didn't think it could be applied until I saw how you first expanded it and integrated into logarithms. After the first integration was explained yesterday, the second integration became obvious, but still, thanks for walking me through the steps you took for both!