Integral that evaluates to a $\tanh^{-1}$, with initial value conditions I'm trying to solve $\displaystyle\int \frac{m \; dv }{mg-kv^2}$. Plugging this into Wolfram Alpha produces $$ \sqrt{\frac{m}{gk}} \tanh^{-1} \left( \sqrt{\frac{k}{gm}}v \right).$$
My question is, what happens when $\displaystyle v> \sqrt{\frac{gm}{k}}$? Then it seems like you're taking the inverse hyperbolic tangent of something greater than $1$, which isn't real.
So is there no real solution to this integral for $\displaystyle v> \sqrt{\frac{gm}{k}}$? Or if there is, what is it?
 A: This formula
$$ \sqrt{\frac{m}{gk}} \tanh^{-1} \left( \sqrt{\frac{k}{gm}}v \right).$$
shows the time $t$ elapse for a mass $m$, falls with $v_0=0$ from height $h$ and has velocity $v$.
For natural phenomenons, we can't decide based on pure mathematical formulas and then need to know all physical parameters which interpret their behaviors.
This value of time is determined by velocity $v$ which specifies by nature and we can't describe it by arbitrary values.
It's clear when $\displaystyle v> \sqrt{\frac{gm}{k}}$, from mathematically rules we may deduce some properties of this experiment, but these properties aren't valuable in physics systems.
A: Let we re-construct the problem from scratch. We have a free-fall motion ($a=g$) and a dissipative force (friction $-kv^2$). Our body starts at $t=0$ with initial speed $v_0=v(0)$ and we want to understand its speed at $t=T>0$. We have:
$$ a(t) = \frac{dv}{dt} = \frac{mg-kv^2}{m} = g-\frac{k}{m}v^2$$
hence
$$ \frac{dt}{dv} = \frac{1}{g-\frac{k}{m}v^2} $$
and by integrating both sides over the interval $(v(0),v(T))$ we get:
$$ T = \int_{v(0)}^{v(T)}\frac{dv}{g-\frac{k}{m}v^2} $$
We may notice that $v_L=\sqrt{\frac{mg}{k}}$ is a limit speed: if $v_0<v_L$, $v(t)$ is an increasing function bounded above by $v_L$; if $v_0>v_L$, $v(t)$ is a decreasing function bounded below by $v_L$. In the first case
$$ T = \sqrt{\frac{m}{gk}}\left[\text{arctanh}\left(\frac{v(T)}{v_L}\right)-\text{arctanh}\left(\frac{v(0)}{v_L}\right)\right]$$
from which:
$$ \exp\left(2T\sqrt{\frac{gk}{m}}\right)=\frac{v_L+v(T)}{v_L-v(T)}\cdot\frac{v_L-v_0}{v_L+v_0} $$
that holds also in the second case$^{(*)}$. Solving that with respect to $v(T)$,
$$ \frac{1+v(T)/v_L}{1-v(T)/v_L}=\frac{1+v_0/v_L}{1-v_0/v_L}\,\exp\left(2T\sqrt{\frac{gk}{m}}\right)$$
leads to:
$$\boxed{\, v(T) = v_L\cdot\frac{\frac{1+v_0/v_L}{1-v_0/v_L}\,\exp\left(2T\sqrt{\frac{gk}{m}}\right)-1}{\frac{1+v_0/v_L}{1-v_0/v_L}\,\exp\left(2T\sqrt{\frac{gk}{m}}\right)+1}.}$$
$(*)$ The crucial fact is that $\frac{v_L-v_0}{v_L-v(T)}$ is positive anyway for $T>0$, and 
$$\exp\left(2\,\text{arctanh}(z)\right) = \frac{1+z}{1-z} $$
holds for every $z\neq 1$. Of course we have not studied the case $v_0=v_L$: in such a case the speed is simply constant.

So, long story short: the solution of the paradox is that the
  non-integrable singularity of $\frac{m}{mg-kv^2}$ is actually never
  enclosed in the real integration range, because $v(0)$ and $v(T)$
  always lie on the same side of $v_L$.

A: Indefinite integration is a useful tool, but it can lead us astray at times, including in cases when the integrand has non-integrable singularities. The proper resolution of this problem is definite integration with symbolic limits. This technique has the bonus of not having to find the integration constant at the end of the calculation. 
In this context, we are looking at:
$$\int_{v_0}^v \frac{m}{mg-ku^2} du.$$
Here the integrand, because it has non-integrable singularities at $u_1:=-m^{1/2} g^{1/2} k^{-1/2},u_2:=m^{1/2} g^{1/2} k^{-1/2}$, tells us that $v_0$ and $v$ must be in the same one of the regions $(-\infty,u_1),(u_1,u_2),(u_2,\infty)$, otherwise the integral does not converge. Let us write the partial fraction expansion:
$$\frac{m}{mg-ku^2} = -\frac{m}{k} \frac{1}{u^2-\frac{mg}{k}} = -\frac{m}{k} \frac{1}{u_2-u_1} \left (  \frac{1}{u-u_2} - \frac{1}{u-u_1} \right ).$$
For $v,v_0 \in (u_1,u_2)$, the end result after cancelling minus signs is that we have two positive terms, so that integration gives us
$$\left. \frac{m}{k} \frac{1}{u_2-u_1} \left [ \ln(u_2-u) + \ln(u-u_1)  \right ] \right |_{u=v_0}^{u=v}.$$
For $v,v_0 \in (u_2,\infty)$ the integration goes differently, resulting in
$$\left. \frac{m}{k} \frac{1}{u_2-u_1} \left [ -\ln(u-u_2) + \ln(u-u_1) \right ] \right |_{u=v_0}^{u=v}.$$
This seems like a minor algebra quirk at a glance, but actually this has a huge impact on the qualitative behavior of the final solution to the differential equation. Physically, when $v_0 \in (u_1,u_2)$, the object will accelerate to its terminal velocity; when $v_0 \not \in [u_1,u_2]$, the object will decelerate to its terminal velocity. (What happens if $v_0=u_1$ or $v_0=u_2$? Why can't this derivation tell us that?)
(Please do check my minus signs, it's quite possible I made a mistake somewhere.)
