Is it possible to integrate .|' v^2dx ?

You say you do not have an expression for acceleration or velocity, but do you have one for position? If you do this is quite a simple integral. Velocity is the first derivative of the position function so v^2 would just be that derivative squared.

example:

s(x) = x^2 +4x + 2 <------------ position function

v(x) = 2x + 4 <-------------- velocity function

v^2 = (2x + 4)^2

∫(2x + 4)^2 dx

u = 2x + 4

du = 2 dx

dx = 1/2 du

∫ u/2 du

u^2/4 + C

(2x + 4)^2/4 + c

well velocity itself is distance per unit time like meters/second. The velocity is the derivative of a position function. Acceleration is the derivative of velocity.

in this case v^2 is treated as a constant because the differential is dx. so the integral would be (v^2)x+c, where c is a constant. You add c to indefinite integrals.