= ∫(x+1)(x−1)(x+1)(x+2)dx
= ∫(x−1)(x+2)dx
= ∫(x+2−2−1)(x+2)dx
= ∫(x+2−3)(x+2)dx
= ∫(x+2)−(3)(x+2)dx
= ∫(x+2)(x+2)dx+ ∫−(3)(x+2)dx
= ∫1dx− ∫3(x+2)dx
= ∫1dx−3 ∫1(x+2)dx
= x−3 ln(x+2) +C