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.