How to evaluate ∮C z/(2z²-3z-2) dz where C is the circle |z|=1 by Cauchy's Integral Formula

How about by direct method and Residue method?

Thanks for the help!!

Note that 2z^2 - 3z - 2 factors as (z-2)(2z+1) or equivalently 2(z-2)(z+(1/2)).
So z/(2z²-3z-2) has simple poles at z = 2 and z = -1/2, but only the pole z = -1/2 is inside C.
Thus,

∮C z/(2z²-3z-2)
= ∮C z/[2(z-2)(z+(1/2))]
= 2pi i {Residue of z/[2(z-2)(z+(1/2))] @ z = -1/2}
= 2pi i (-1/2) / [2((-1/2) - 2)]
= 2pi i (-1/2) / (-5)
= pi i/5

Lord bless you today!