Evaluate $\int _{|z|=2}ze^{\frac{3}{z}} dz$ How to integrate this 

$\int _{|z|=2}ze^{\frac{3}{z}} dz$

I tried to do this by substituting $z=2e^{i\phi};0\leq \phi\leq 2\pi$

But cannot proceed further.Any help
$$\int_{|z|=2}ze^{3/z}\,dz=-2\pi iRes(f,\infty)=2\pi iRes\left(\frac{1}{z^2}f(1/z),0\right)$$ where , $f(z)=ze^{3/z}$.

Now , $\frac{1}{z^2}f(1/z)=\frac{1}{z^3}e^{3z}$

So, $Res\left(\frac{1}{z^2}f(1/z),0\right)=Res\left(\frac{1}{z^3}e^{3z},0\right)=\frac{1}{2!}\lim_{z\to 0}\frac{d^2}{dz^2}e^{3z}=\frac{9}{2}$

So, required answer will be $9\pi i$.
Expanding and integrating term by term we end up with integrating 9/(2z) round the circle given. The answer is 2i(pi)(9/2)=9i(pi).