Cauchy Integral for complex analysis

Question

I need to find the below without residue theorem and only using Cauchy Integral theorem. $$\int\frac{\cos z}{z^2(z-2)}dz$$

The problem I am having right now is how to express $\frac{\cos z}{z^2(z-2)}$ in terms of partial fraction decomposition. So that I can apply the Cauchy Integral formula of $$f^{(n)}(z) = \frac{n!}{2i\pi}\int\frac{f(w)}{(w-z)^{n+1}}\,dw.$$

Answer 1

$$\dfrac{1}{z^2(z-2)}=\frac{a}{z-2}+\frac{b}{z}+\frac{c}{z^2}=\dfrac{(a+b)z^2+(c-2b)z-2c}{z^2(z-2)}$$

Hence $c=-\dfrac12,b=-\dfrac14,a=\dfrac14$.

Answer 2

You can use the fact that$$\frac{\cos z}{z^2(z-2)}=-\frac{\cos z}{4z}-\frac{\cos z}{2z^2}+\frac{\cos z}{z-2}.$$