Confusion about a way to compute a residue at a pole Suppose I have a function of the form $$f(x)=\frac{1}{(x-a)(x-b)^2(x-c)^3}$$
Clearly, I have a simple pole at $a$, and poles of order 2,3 at $b,c$, respectively. 
By definition, the residue at $x=\alpha$ is the coeffecient of the term $(x-\alpha)^{-1}$ in the Laurent expansion of $f(x)$ centered at $x=\alpha$. From my experience, this residue at $x=\alpha$ is equivalent to the coefficient of the term $(x-\alpha)^{-1}$ in the partial fraction decomposition of $f(x)$. However, the partial fraction of $f(x)$ will not be a Laurent series since it has no center (I.e., there are denominators of the form $x-\beta$ with $\beta \not = \alpha$). Is there a rule that says that the coefficient of $(x-\alpha)^{-1}$ in the partial fraction decompositon is the residue at $x=\alpha$, even if the partial fraction decompoition is not a Laurent expansion?

Yes, there is such a rule. If you have a partial fraction decomposition
$$f(z) = h_\alpha(z) + h_\beta(z) + h_\gamma(z) + g(z),$$
where $h_w$ is the principal part of $f$ in $w$, and $g$ is the remaining holomorphic part of $f$, then you can develop $h_\beta,\, h_\gamma$, and $g$ in Taylor series about $\alpha$ to obtain the Laurent series of $f$ in the annulus $0 < \lvert z-\alpha < d$, where $d$ is the distance to the nearest other singularity ot the boundary of the domain of $f$, whatever is smaller.
The point is that everything but the fractions $\frac{c_k}{(z-\alpha)^k}$ in the partial fraction decomposition of $f$ is holomorphic in a neighbourhood of $\alpha$, so can be expanded into a power series around $\alpha$ and therefore doesn't influence the coefficient of $(z-\alpha)^{-1}$.
