Need to integrate 1/(x^p) for any constant p. It is [x^(-p+1)]/(-p+1) for all p != -1, what about for r = -1?

Question

oops, meant for all r!=1

Answer 1

Actually, the formula you have is correct for all p != +1. (if p=-1, then your integrand simplifies to just x, which is of course (x^2)/2)

For p = 1, the integral of 1/x is ln(x) (natural log of x).

Answer 2

The integral of y =1/x is LnIxI