Let's use the Cauchy Residue Theorem to solve this problem.
First, we need to find the poles of f(z).
f(z) has poles at z = i, -i, -1.
Let's consider the semicircle CR of radius R with center at the origin, parameterized by z(t) = Re^(it) with t in [0, π].
Using the Cauchy Residue Theorem, we have:
∫CR f(z) dz = 2πi Res(f, i) + 2πi Res(f, -i) + 2πi Res(f, -1)
Let's find the residues of f(z) at each of the poles.
At z = i, we have:
Res(f, i) = lim(z→i) (z-i) f(z)
= lim(z→i) (z-i) [(z²+2) (z²+3) (z²+2z+1)]
= (i²+2) (i²+3) (i²+2i+1)
= -4i
At z = -i, we have:
Res(f, -i) = lim(z→-i) (z+i) f(z)
= lim(z→-i) (z+i) [(z²+2) (z²+3) (z²+2z+1)]
= (-i²+2) (-i²+3) (-i²-2i+1)
= 4i
At z = -1, we have:
Res(f, -1) = lim(z→-1) (z+1) f(z)
= lim(z→-1) (z+1) [(z²+2) (z²+3) (z²+2z+1)]
= (1²+2) (1²+3) (1²-2+1)
= 60
Substituting these values into the Cauchy Residue Theorem, we get:
∫CR f(z) dz = 2πi (-4i) + 2πi (4i) + 2πi (60)
= 120π
Therefore, lim(R→∞) ∫CR f(z) dz = 120π.