Ah, then indeed you can use the integral you gave.
An alternative is to note that 3V + Vx = 0 is a linear ordinary differential equation. In that case, you can usually stick in [itex]V(x) = e^{\lambda x}[/itex] as a "trial" solution. The differential equation then turns into a polynomial equation for [itex]\lambda[/itex] with solutions [itex]\lambda_1, \cdots, \lambda_n[/itex], where n is the order of the ODE (in this case, n = 1). The general solution to the ODE is then
[tex]V(x) = A_1 e^{\lambda_1 x} + \cdots + A_n e^{\lambda_n x}[/itex]<br />
where the A<sub>i[/i] are to be determined (for example from initial or boundary values).</sub>[/tex]