$$m\dot{v}=-cv^2-f_{fr}$$
In a textbook that I am reading, I am tasked to solve the above differential equation, providing $t$ as an explicit function of $v$.
Here is my attempt at solving this differential equation:
$m\dot{v}=-cv^2-f_{fr}$
$\implies \frac{m\dot{v}}{-cv^2-f_{fr}} = 1$
$\implies -m \int_{v_0}^{v} \frac{1}{cv^2+f_{fr}} dv = t$
$\implies t = \frac{-m}{f_{fr}} \int_{v_0}^{v} \frac{1}{1+\frac{c}{f_{fr}}v^2} dv$
$\implies t = \frac{-m}{\sqrt{cf_{fr}}} \arctan \left( \sqrt{\frac{c}{f_{fr}}}v \right) _{v_0}^v$
$\implies t = \frac{m}{\sqrt{cf_{fr}}} \arctan \left( \sqrt{\frac{c}{f_{fr}}}(v_0-v) \right)$
However, the answer provided is $t = \frac{m}{\sqrt{cf_{fr}}} \left( \arctan \sqrt{\frac{c}{f_{fr}}} v_0 - \arctan \sqrt{\frac{c}{f_{fr}}} v \right)$. To my knowledge, these two answers are not equivalent. Is there any error in my working that led to a different answer?
Nevermind, I skipped too many steps.
$\implies t = \frac{-m}{\sqrt{cf_{fr}}} \arctan \left( \sqrt{\frac{c}{f_{fr}}}v \right) _{v_0}^v$
should be written as
$\implies t = \frac{-m}{\sqrt{cf_{fr}}} \left[ \arctan \left( \sqrt{\frac{c}{f_{fr}}}v \right) \right]_{v_0}^v$
so that the definite integral can be evaluated without mistakes