Not quite
int(dx/3x) is not ln(3x) but 1/3ln(3x) +c = 1/3ln(x) + c (1/3ln(3)is absorbed into c
To see this differentiate ln(3x) -- you don't get 1/3x
When integrating int(dx/3x) if you leave the 3 in the integral
you should use substitution u = 3x du = 3dx
1/3du = dx and you get 1/3int(1/udu)