How to calculate $\int_{\partial D(0;a)} \frac{dz}{(z-a)(z-b)}$

Question

How to calculate $\displaystyle\int_{\partial D(0;r)} \frac{dz}{(z-a)(z-b)}$ when $|a|<r<|b|$?.

My first idea: I tried to separate by partial fractions, so for some $A,B$ \begin{equation} \int_{\partial D(0;r)} \frac{dz}{(z-a)(z-b)} = \int_{\partial D(0;r)} \frac{Adz}{z-a} + \int_{\partial D(0;r)} \frac{Bdz}{z-b}, \end{equation} and, by Cauchy integral formula, considering $f\equiv 1$ (is holomorphic) \begin{equation} \int_{\partial D(0;r)} \frac{Adz}{z-a} = A2\pi i, \end{equation} and \begin{equation} \int_{\partial D(0;r)} \frac{Bdz}{z-b} = B2\pi i. \end{equation} with which we get what we want. Is this procedure correct?

My second idea: Again reasoning by the Cauchy integral formula, considering $f(z) = \frac{1}{z-b}$ holomorphic on $D(0;r)$, then \begin{equation} \int_{\partial D(0;r)} \frac{dz}{(z-a)(z-b)} = 2\pi i \frac{1}{a-b}. \end{equation} Is this procedure correct?

Regards!

Answer 1

The second method is correct and it leads to the right answer. The first method also works, if you use the fact that$$\frac1{(z-a)(z-b)}=\frac1{a-b}\left(\frac1{z-a}-\frac1{z-b}\right)$$together with the fact that$$\oint_{\partial D(0,r)}\frac1{z-a}\,\mathrm dz=1\quad\text{ and }\quad\oint_{\partial D(0,r)}\frac1{z-b}\,\mathrm dz=0.$$