Given the velocity y = Integral (dt/t).

To find the distance function.

Le s(t) be the distance travelled in time t.

Then velocity = ds/dt = Int (1/t)

Therefore ds = Int (dt/t )

We integrate both sides to get the distance function s(t) :

Therefore s(t) = logt +C.

Therefore the distance function s(t) travelled in time t is given by:

s(t) = logt +C, where C is the constant of integration which could be determined by the initial conditions.

The velocity is the derivative of distance, with respect to time:

v = ds/dt

vdt = ds

We'll integrate both sides:

Int vdt = Int ds

Int ln t dt/t = s

We notice that if we'll re-write the function, we'll have:

If we'll substitute ln t = u and we'll differentiate, we'll get:

dt/t = du

We'll re-write the integral, changing the variable:

Int ln t dt/t = Int u du

Int u du = u^2/2 + C

We'll substitute u by ln t:

Int ln t dt/t = (ln t)^2/2 + C

**The function of distance is: d(t) = (ln t)^2/2.**