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 $0$, $t$. 
 A: As suggested by Arturo:
$$ \begin{eqnarray}
     \int_{0}^{t} \frac{\dot{v}(t)}{a v(t)^2+ b v(t) + c} \mathrm{d} t &=& \int_0^t \mathrm{d} t \\
     \int_{v_0}^{v_t} \frac{\mathrm{d} u}{a u^2 + b u + c} &=& t \\
     \int_{v_0}^{v_t} \frac{4 a }{\left(2 a u + b\right)^2 + 4 a c - b^2}  \mathrm{d} u &=& t \\
     \left.\frac{2}{\sqrt{4 a c - b^2}} \arctan\left(\frac{2 a u + b}{\sqrt{4 c a - b^2}}\right)\right|_{v_0}^{v_t} &=& t \\
     \frac{2}{\sqrt{4 a c - b^2}} \arctan\left(\frac{2 a v_t + b}{\sqrt{4 c a - b^2}}\right) &=& \frac{2}{\sqrt{4 a c - b^2}} \arctan\left(\frac{2 a v_0 + b}{\sqrt{4 c a - b^2}}\right) + t \\
    \frac{2 a v_t + b}{\sqrt{4 c a - b^2}} &=& \tan\left( \arctan\left(\frac{2 a v_0 + b}{\sqrt{4 c a - b^2}}\right) + \frac{\sqrt{4  c a - b^2}}{2} t \right) \\
    \frac{2 a v_t + b}{\sqrt{4 c a - b^2}} &=& \frac{ \frac{2 a v_0 + b}{\sqrt{4 c a - b^2}} + \tan\left(\frac{\sqrt{4  c a - b^2}}{2} t\right)}{1 - \frac{2 a v_0 + b}{\sqrt{4 c a - b^2}} \tan\left(\frac{\sqrt{4  c a - b^2}}{2} t\right)}
\end{eqnarray}
$$
Now one solves for $v_t$, and integrates over time:
$$
     x(t) = x(0) + \int_0^t v_t \mathrm{d} t
$$
The integral is not likely to admit evaluation in closed form. Typically trajectories for such system are defined parametrically.
Added
As pointed out in @ChristianBlatter answer, the integral is indeed elementary:
$$
   x(t) = x(0) - \frac{b}{2a} t - \frac{1}{2a} \log\left(\frac{4 a \left(a v_0^2 + b v_0 + c\right)}{4 a c-b^2} \cos^2\left(\frac{t}{2} \sqrt{4 a c - b^2} + \arctan\left(\frac{2 a v_0 + b }{\sqrt{4 a c - b^2}}\right)\right)\right)
$$
Here is a comparison of the exact solution to the numerical solution of the differential equation in Mathematica:

Several physical conclusions:


*

*if $4 a c - b^2 < 0$, trigonometric cosine becomes hyperbolic cosine, and position $x(t)$ increases with time linearly, approximately linearly.

*if $4 a c - b^2 > 0$, the position $x(t)$ plunges to $-\infty$ in finite time, determined by zero of the cosine function.

*if $4 a c = b^2$, the solution is easy to re-derived. It also goes to $-\infty$ in finite time (if $2a v_0 + b > 0$), or goes $\pm\infty$ (if $2 a v_0 + b < 0$), depending on initial conditions:
$$
  x(t) = x(0) - \frac{ b t}{2a} - \frac{1}{a} \log\left( 1 - \frac{b t}{2} - a t v_0\right)
$$

A: "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)
$$x(t)=\int_{t_0}^t v(t)\ dt=\int _{t_0}^t \tan t\ dt=-\log(\cos t)\Bigr|_{t_0}^t\ .$$
