I am trying to compute the integral $\int_{\gamma} \frac{1}{(z-a)(z-b)} dz$ without any residue theorems or Cauchy's theorem, where $\gamma$ is a circled center at $0$ of radius $r$ with $|a| < r < |b|$. Using PFD, the integral can be rewritten as
$$\frac{1}{a-b} \left[ \int_{\gamma} \frac{1}{z-a} dz - \int_{\gamma} \frac{1}{z-b} dz \right].$$
If $\gamma$ is parametrized as $\gamma(t) = re^{it}$, then the first integral becomes $$\int_{0}^{2 \pi} \frac{ i r e^{i t}}{re^{it} - a} dt.$$
I'm not sure how to deal with that integral. It looks like the integral of $u'/u$, but that would involve logs and one has to be careful with branch cuts....
For $a\in[0,\infty)$ and all $t\in (0,2\pi)$, $re^{it}-a\in\mathbb C\setminus[0,\infty)$, so we can take the branch cut to be at the positive real axis and the integral is $\lim_{t\to 2\pi^-}\log(re^{it}-a)-\lim_{t\to 0^+}\log(re^{it}-a)=2\pi i$. The general case is similar, simply that the branch cut is along a different line.