Specific formulation of a Laurent Series?

Question

I'm rather new to this branch of calculus, but I've been trying to learn about it on my own time. I currently rather fluently understand more basic topics such as Taylor Series, which come with an easy little formula for determining them$\left(T_a(f(x))=\sum_{k\geq0}\frac{f^k(a)}{k!}(x-a)^k\right)$.

This is how I currently go about finding a Laurent Series, and I'm pretty sure it's wrong. Let's look at the example $$f(z)=\frac{1}{z+2}=-\frac1{x_1}=\frac1{x_2}:x_1=-(z+2)\wedge x_2=(z+2)$$ Basically, using these x-substituted forms, I take two Taylor Series combined to get one who has indices on $(-\infty,\infty)$. So: $$-\frac1{x_1}=-x_1^{-1}=-T_1(x_1^{-1})=-\sum_{k=0}^{\infty}(-1)^k(x_1-1)^{k-1}=\sum_{k=0}^{-\infty}(-1)^k(x_1-1)^{k+1}\\ \frac1{x_2}=x_2^{-1}=T_1(x_2^{-1})=\sum_{k=0}^{\infty}(-1)^k(x_2-1)^{k-1}\\ \therefore \sum_{k=0}^{-\infty}(-1)^k(x_1-1)^{k+1}+\sum_{k=0}^{\infty}(-1)^k(x_2-1)^{k-1}\\ \downarrow\text{substituting }x_1, x_2 \downarrow\\\sum_{k=0}^{-\infty}(-1)^k(-z-3)^{k+1}+\sum_{k=0}^{\infty}(-1)^k(z+1)^{k-1}\\ \sum_{k=-\infty}^{\infty}2(-1)^k(-z-3)^{k+1}+(z+1)^{k-1} $$ So yeah, even assuming this is the right sum equation, I'm still hella confused. Firstly, my end goal with understanding Laurent series is being able to calculate contour integrals, which you must use the coefficient of the $-1^{\text{st}}$ index of the series to calculate the residue$\left(\oint_{\gamma} f(z) dz=2\pi i\sum_{k\leq n}\text{Res}\left(f\left(z\right),z_k\right)\right)$. If someone could help explain to me what the correct way to calculate this is in the context of Taylor Series or introductory complex analysis terminology, that would be awesome. Thanks in advance!