What is the $x(t)$ function of $\dot{v} = a v² + bv + c$ to obtain $x(t)$ How to solve $$\frac{dv}{dt} = av^2 + bv + c$$ to obtain $x(t)$, where $a$, $b$ and $c$ are constants, $v$ is velocity, $t$ is time and $x$ is position. Boundaries for the first integral are $v_0$, $v_t$ and $0$, $t$ and boundaries for the second integral are $0$, $x_{max}$ and $0$, $t$. 

"Separating variables" means writing
$${dv\over av^2+bv +c}=dt\ .\qquad(1)$$
The next step depends on the values of $a$, $b$, $c$. Assuming $a>0$ one has
$$a v^2+bv +c={1\over a}\Bigl(\bigl(av +{b\over 2}\bigr)^2+{4ac -b^2\over 4}\Bigr)\ ,$$
so that after a linear substitution of the dependent variable $v$ the equation $(1)$ transforms into   one of
$${du\over 1+u^2}= p\ dt,\quad{du\over u^2}=p\ dt,\quad {du\over u^2-1}=p \ dt\ ,$$
depending on the value of $4ac-b^2$. Up to scalings and shifts the first form implies
$$\arctan u =  t\quad{\rm or}\quad  u=\tan  t\ .$$
It follows that (again neglecting scalings and shifts)
$$x(t)=\int_{t_0}^t v(t)\ dt=\int _{t_0}^t \tan t\ dt=-\log(\cos t)\Bigr|_{t_0}^t\ .$$
