Because there is only one third-order pole z= 1 inside the circle, the remainder at z= 1 is found first. According to the calculation rules of residue, Res[f(z), 1]=

(1/2) limd2 (2z3-3z2+4z+1)/dz2 (z tends to 1)=3, so the integral = 2π IRES [f (z), 1].