The function has poles of order one at z=0 and z=+2 and a pole of order two at z=+1.
According to the Residue Theorem, the integral is equal to the sum of all residues of your function. Pay attention that all poles must be enclosed by the circle |z|=5/2. In this case, all poles are enclosed by that circle.
So we start by evaluating the residue at z=0 (simple pole) which in this case is the limit as z tends to zero of your given function where you remove the term z at the denominator. In this case you have z+3/((z-2)(z-1)^2) and if you substitute z=0 you obtain that the residue is -3/2
Secondly the residue at z=2 is the limit as z tends to 2 of your given function where you remove the term (z-2) at the denominator. In this case you have (z+3)/(z(z-1)^2) and if you substitute z=2 you obtain that the residue is 5/2
Then we need to evaluate the residue at z=1, but in this case you have a pole of order two, so you have to consider your original function, remove the term (z-1)^2 at the denominator, take the derivative of the resulting function and then evaluate the limit as z to 1.
If you remove the term (z-1)^2 an then you take the derivative you get z(z-2)-(z+3)(2z-2)/(z^2(z-2)^2)^2 and now you consider the limit as z to 1 which gives you 1. So finally the residue at z=1 is equal to 1.
The integral is therefore the sum of the residues (multiplies by 2 pi j, where j is the imaginary unit). In this case the sum is equal to zero, therefore the contour integral is equal to zero. -3/2 -1 + 5/2 =0
A faster way would have been to note that the function is regular at z=infinity, and hence the only poles are at z finite and they are the three which we already mentioned. Hence, since the sum of all residues( z=finite +z=infinity) must be equal to zero we could have concluded that the integral is zero without evaluating the residues.