Phew that's a lot of work

The work done by a force F on a particle moving from position x1 to x2 on a straight line is given by $\int_{x_{1}}^{x_{2}} F dx$

If the motion of the particle is described by the expression F = K/v, then it is also true that Fv = K and F $\frac{dx}{dt}$ = K

$$

Integrate both sides with respect to t to obtain $$ $\int_{0}^{t} F \frac{dx}{dt} dt$ = $\int_{0}^{t} K dt$ $$ which becomes: $$ $\int_{x(0)}^{x(t)} F dx$ = $\int_{0}^{t} K dt$

where x(0) and x(t) correspond to x1 and x2 respectively.

The left side of the expression represents the work done on the particle from time t to time 0, and the right side evaluates to Kt.