One could use the residue theorem to find the residues of those simple poles, but calculating the integral directly would probably be very difficult - the easiest way to calculate the relevant integrals would be to use the residue theorem.
It's easy to find the residues directly and the point of this exercise is to practice doing so. Either expand the terms you produced after partial fractions via geometric series to produce a Laurent series and read off the $a_{-1}$ coefficient, or use the short cut limit method for simple poles: $\displaystyle\text{Res}_{z=a} f(z) = \lim_{z\to a} (z-a) f(z) .$
I'll show you an example so you can get started. Using the limit method is certainly the quickest, but it is good practice to find Laurent series anyway. Say we want to find the residue at $z=0$. We want to write a Laurent series $\displaystyle f(z) = \sum_{n=-\infty}^{\infty} a_n z^n $ valid in some annulus around $z=0$. We know $$ f(z) = \frac{1}{2(z-2)} + \frac{1}{1-z} + \frac{1}{2z}.$$ The last term is already in the form we want. The middle term expands to $$ \frac{1}{1-z} = \sum_{n=0}^{\infty} z^n$$ for $|z|<1 $ and the first term can be written $$ \frac{-1/4}{1-z/2} = \frac{-1}{4} \sum_{n=0}^{\infty} \left( \frac{z}{2} \right)^n $$ for $|z/2|<1, \text{ or } |z|<2. $
Thus, in the annulus $0<|z|<1$ we have the Laurent series $$f(z) = \frac{1}{2z} + \sum_{n=0}^{\infty} \left( 1 - \frac{1}{2^{n+2}} \right) z^n.$$
We can read off that the residues at $z=0$ is $1/2$. Of course, we could have also noticed that the partial fraction decomposition left us with the form $$ f(z) = \frac{1}{2z} + \text{Function analytic at z=0} $$ and so the only negative powered term will be $\displaystyle \frac{1}{2z}$.
An important point is that a "pole" is actually the same thing as a removable singularity, if we think of our function as a map that takes values on the Riemann sphere (which is the complex plane with a point at $\infty$ added; the complex structure near $\infty$ comes from the map $z\mapsto 1/z$).
So a function that has a removable singularity or a pole at $z_0$ doesn't have a "real" singularity there at all; rather, we can extend the function to an analytic or meromorphic function in $z_0$. If we cannot extend the function in this way, the singularity is indeed "essential"; i.e., we cannot get rid of it. Thus the terminology is not one that is merely used for convenience or pedagogical purposes; rather, it is extremely natural.
As has already been mentioned, the magic of complex numbers results in many beautiful facts about essential singularities: functions with these singularities are very far from extending continuously.
The simplest of these facts is the Casorati-Weierstraß Theorem: The image of a neighborhood of an essential singularity is dense in the complex plane.
This is just a consequence of the removable singularities theorem. (If $f$ omitted a neighborhood of $a$, we could postcompose $f$ with a Möbius transformation that takes $a$ to infinity and see that the resulting function has a removable singularity.)
The most well-known result of this type is Picard's theorem which was already mentioned.
There are various beautiful strengthenings of Picard's theorem that arise from Nevanlinna theory, and Ahlfors's theory of covering surfaces.
So all essential singularities have some things in common, but on the other hand this should not lead us to believe that they are all the same. What they have in common is complicated behaviour, but they can be complicated in very different ways! Indeed, different transcendental entire functions (those that have an essential singularity at infinity; i.e. are not polynomials) can vary very much with respect to their behavior near infinity. Just for example, for some such functions, such as $z\mapsto e^z$, there exist curves tending to infinity on which the function is bounded, while for others this is not the case.
Best Answer
One must be careful here. We can define $\log(z)$ for $z\in\mathbb{C}$ if we leave out the negative real axis. It is this branch cut that causes problems.
Riemann's Theorem on removable singularities says that if $f$ is defined on a punctured neighborhood of $a$ and $$ \lim_{z\to a}(z-a)f(z)=0 $$ then $f$ has a removable singularity at $a$. Indeed, if we consider $$ \left|z\log(z)\right|\le\left|z\right|(\left|\log|z|\right|+\pi) $$ and $$ \lim_{t\to0^+}t\log(t)=0 $$ we get that $$ \lim_{z\to0}z\log(z)=0 $$ for all $z\in\mathbb{C}$ as long as $\boldsymbol{z}$ is not on the negative real axis. The fact that $z\log(z)$ is not defined on a full punctured neighborhood of $0$, means that although we have the limit above, Riemann's Theorem does not apply.
This is the problem with $\log\left(1-\frac1z\right)$; we cannot define this function in a punctured neighborhood of $0$ or $1$. These are branch points of this function.