Integral along the boundary of a triangle of $\int_{\partial\bigtriangleup}\frac{f(z)dz}{z(z^2-1)}$

Question

Given a function $f$ holomorphic in a closed triangle $\bigtriangleup$ I need to find the integral $$\int_{\partial\bigtriangleup}\frac{f(z)dz}{z(z^2-1)}$$ in every case in which $-1,0,1$ doesn't belong to the boundary of the triangle. I know that if neither of the three points belongs to the triangle, then the function that I'm integrating is holomorphic in convex open set that contains the triangle and , by Cauchy's theorem , the value of the integral is $0$. However , I don't know what to do in any of the other cases. Any ideas?

Answer 1

Hint: Make use of the partial fraction decomposition $$\frac{1}{z(z^2-1)} = -\frac{1}{z}+\frac{1}{2 (z+1)}+\frac{1}{2 (z-1)}, $$ and Cauchy's integral formula.