AMC 12 Mock Exam 5 Problem 21

Let $P(x)=x^5-1=\prod_{k=0}^{4}(x-\omega^k)$. Differentiating a product (or comparing coefficients after expanding), $\frac{P'(x)}{P(x)}=\sum_{k=0}^{4}\frac{1}{x-\omega^k}$. Without calculus: $\sum_k\frac{1}{x-\omega^k}=\frac{\sum_k\prod_{j\ne k}(x-\omega^j)}{P(x)}$, and the numerator is the polynomial $5x^4$ (it has degree $4$, leading coefficient $5$, and the coefficient of $x^{4-m}$ is $(5-m)$ times the corresponding elementary symmetric sum, which vanishes for $1\le m\le4$).

At $x=2$: $\sum_{k=0}^{4}\frac{1}{2-\omega^k}=\frac{5\cdot16}{31}=\frac{80}{31}$. Removing the $k=0$ term $\frac{1}{2-1}=1$ leaves $\frac{80}{31}-1=\boxed{\tfrac{49}{31}}$.