consider an object with a mass $m$ falling in a fluid with a drag force proportional to its velocity squared $(f=kv^2)$. the governing differential equation can be found using Newton's second law of motion as $$ A \frac{dv}{dt} + v^2 = v_{lim}^2 ~,~v(0)=v_0$$ where $A=\frac{m}{k}$ and $v_{lim}=\sqrt{\frac{mg}{k}}$ is the final velocity with the initial condition $v(0)=v_0$.
the solution of the equation is $$v(t)=v_{lim} \tanh \left(\frac{v_{lim}t}{A}+ \tanh^{-1} \left(\frac{v_0}{v_{lim}} \right) \right)$$ where $\tanh^{-1}$ is the inverse hyperbolic tangent.

The given solution does reflect the physical phenomenon in the case of $(v_0<v_{lim})$ ie the velocity increases from $v_0$ to $v_{lim}$.

In the case of $(v_0>v_{lim})$, the expected physics "behavior" of the solution is that the velocity decreases over time (from $v_0$ to $v_{lim}$) and yet we never see such decrease when plotting the function due to the fact that the $\tanh^{-1}(x)$ function is only defined for $x<1~ie~(v_0<v_{lim})$.
So, is there maybe another formula that better describe the second case?