are you sure you have the limits right? usually they put the smaller number at the bottom. should the -2 be 2?

int{-2:0}[(x + 2)^2]dx

we proceed by substitution

let u = x + 2

=> du = dx

so our integral becomes

int{u^2}du

= (1/3)u^3 + C

= (1/3)(x + 2)^3 + C

evaluating between -2 and 0 we get

int{-2:0}[(x + 2)^2]dx = (1/3)(0 + 2)^3 - (1/3)(-2 + 2)^3

................................= 8/3