How to find the integral of this problem

|(The top bound is 0 and bottom bound is -1) x^2(x^3 + 2) ^ -2 dx
I'm lost after I get to | x^2 (u)^ -2 du/3x^2

Substitution u = x³ + 2 and therefore du = 3x²dx we get or dx = du/3x². The integral than is
integral(x²*u⁻²*du/3x²) = 1/3*integral(u⁻²du) = -1/3*u⁻¹>>-1/3*(0 -(-1)⁻¹) = -1/3