The problem is the same as in this thread : http://www.mathhelpforum.com/math-he...l-formula.html, with the exception of the integrand : $\displaystyle \frac{\cosh z}{z^4}$.

My attempt : I see the singularity at $\displaystyle z=0$.

So $\displaystyle \int _C \frac{\cosh z}{z^4}dz=2\pi i f(0)$ with $\displaystyle f(z)=\cosh z$.

As $\displaystyle \cosh 0=1$, the integral is worth $\displaystyle 2 \pi i$.

Edit : If the integrand is $\displaystyle \frac{z}{2z+1}$, I get the result $\displaystyle -\pi i$. Am I right?