Question about integral on the complex plane That is the infernal integral:

$\int_{|z|=2}{\frac{e^{z+\frac{1}{z}}}{1-z^2}}dz$

I tried to solve it with residuals, but I can't find the expression of the function in its Laurent form.
Sorry if the question is already posted, i don't speak the language well yet.
 A: $\def\res{\operatorname{res}}$
A simple analysis shows that the function
$$f(z)=\frac{e^{z+\frac{1}{z}}}{1-z^2}
$$
has four singularities: $-1,1,0,\infty$.
Since the function $f(z)$ is everywhere else holomorphic the sum of the residues at these points is zero:
$$
\res(f(z),-1)+\res(f(z),1)+\res(f(z),0)+\res(f(z),\infty)=0.\tag1
$$
The important observation is:
$$
\res(f(z),\infty)=-\res\left(z^{-2}f(z^{-1}),0\right)
=-\res\left(\frac{e^{\frac{1}{z}+z}}{z^2-1}\right)=\res(f(z),0).\tag2
$$
Combining (1) and (2) one easily obtains the final result:

$$\frac1{2\pi i}\oint_{|z|=2}f(z)\,dz=-\res(f(z),\infty)=\frac{\res(f(z),-1)+\res(f(z),1)}2=-\frac{\sinh(2)}2.$$

A: Okay, let me try to find the residue of $e^{z + \frac{1}{z}}$ in $z=0$ (outside of $0$, it is holomorphic). The other poles were treated in the comment already.
First comes the Taylor expansion of $e^x$:
$$
e^{z + \frac{1}{z}} = \sum_{n=0}^\infty \frac{1}{n!} \Big( z + \frac{1}{z} \Big)^n
$$
Using the binomial theorem, one gets
$$
 \Big( z + \frac{1}{z} \Big)^n = \sum_{k=0}^n \frac{n!}{k!(n-k)!} z^{n-k} (\frac{1}{z})^k = \sum_{k=0}^n \frac{n!}{k!(n-k)!} z^{n-2k}
$$
and the series becomes
$$
e^{z + \frac{1}{z}} = \sum_{n=0}^\infty  \sum_{k=0}^n \frac{1}{k!(n-k)!} z^{n-2k}
$$
To write this as a proper Laurent series in $z^k, k=-\infty...\infty$ may be difficult. For the residue, however, we only need to sum all coefficients with  $z^{-1}$, so all terms where $n - 2k = -1$ or $n = 2k-1$.
For each odd $n \geq 0$, there is exactly one $k$ such that $n= 2k-1$. We can turn this around and say for each $k\geq 1$ there is one $n = 2k-1$ such that $z^{n-2k} = z^{-1}$.
For the residue, we sum over all of the coefficients:
$$
\text{Res}[e^{z + \frac{1}{z}} ,0] = \sum_{\substack{k=1 \\ n=2k-1}}^\infty \frac{1}{k!(n-k)!}  = \sum_{k=1}^\infty \frac{1}{k!(k-1)!} = I_1(2) = 1.5906...
$$
where $I_1$ is the modified Bessel function.
I hope this is correct, let me know if I made any mistakes!
EDIT:
Seeing that the comment does not seem to properly adress the question, I add the treatment of the remaining poles:
$$
\frac{e^{z + \frac{1}{z}}}{1-z^2} = \frac{e^{z + \frac{1}{z}}}{(1+z)(1-z)}
$$
has three poles at $z=0, \pm 1$. The pole at $z= 0$ gives residue $\text{Res}\Big[\frac{e^{z + \frac{1}{z}}}{1-z^2}, 0\Big] = \text{Res}\Big[e^{z + \frac{1}{z}}, 0\Big] = I_1(2)$, as $\frac{1}{1-z^2}|_{z=0} = 1$.
The poles at $z= \pm 1$ yield residues:
$$
\text{Res}\Big[\frac{e^{z + \frac{1}{z}}}{1-z^2}, \pm 1\Big] = \mp \frac{e^{\pm 2}}{2} ~~,~~ 
$$
The integral then $\int_{|z|=2} f(z) = 2\pi i \sum_{z_0 \in \{\pm 1,0\}} \text{Res}[f(z), z_0]$ as we know from the residue theorem.
