result of this integral? $\int_{|z|=3} \frac{e^z}{(z-1)(z-2)}$

Question

I think I can solve it by taking: $z(t)=3(\cos t + i \sin t)=3e^{it}$ and by using

$$\int_c f(z)dz = \int_c \frac{dz}{dt}\cdot f(z(t))$$ we have:

$\displaystyle\int_{|z|=3} \frac{e^{3e^{it}}}{(3e^{it}-1)(3e^{it}-2)}\cdot3ie^{it}$ but I can't continue this approach to find a solution without using residue theorem

Answer 1

Hint

$$\frac{1}{(z-1)(z-2)}=\frac{1}{z-2}-\frac{1}{z-1}$$

Thus $$\int_{|z|=3} \frac{e^z}{(z-1)(z-2)}dz=\int_{|z|=3} \frac{e^z}{z-2}dz-\int_{|z|=3} \frac{e^z}{z-1}dz$$

The answer follows easily from Cauchy Integral Formula.