The easiest way to do this is to bound it from below by a function whose integral you already know diverges. For example:
1/(3x^2+6)^(1/3) 1/(9x^2)^(1/3)
since 3x^2+6 9x^2 for all x 1. Then the function on the right is a constant times:
1/x^(2/3)
We know when the integral of 1/x^p from 1 to infinity converges and when it diverges--it converges for p > 1 and diverges for p 1. Since 2/3 1, the integral of 1/x^(2/3) diverges, and hence so do the integrals of the functions above.