$\int\limits_{\gamma} \frac{1}{z-1}$
$\gamma = \{z : \lvert z \rvert = 1\}$
I use Cauchy's integral formula, which says $$\int\limits_{\gamma} \frac{f(z)}{(z-a)^{n+1}} = \frac{2\pi i}{n!} f^{(n)}(a)$$ and I choose $f(z) = 1$ to be my holomorphic function.
$n = 0$ and $a = 1$ (which doesn't matter since my $f(z)$ is a constant function), so evaluating Cauchy's integral formula, $$\int\limits_{\gamma} \frac{1}{z-1} = 2\pi i.$$
Why is this not the correct solution?
If there is a pole on the contour, the integral doesn't converge. If it is a simple pole, you can find the Cauchy principal value (basically, if the integrand has a singularity at $c \in [a,b]$, the CPV is $$ \lim_{\varepsilon \to 0} \int_a^{c-\varepsilon}+\int_{c+\varepsilon}^b) $$ by taking half the residue at $c$, which you prove by deforming the contour, adding a small semicircle around $c$.
For any other sort of pole, the situation can't be tackled in general.