$\int dx/(x^{10000}-1)$ Is there any way to evaluate this indefinite integral using pencil and paper? A closed-form solution exists, because $1/(x^{10000}-1)$ can be expressed as a partial fraction decomposition of the form $\sum c_m/(x-a_m)$, where the $a_m$ are the 10,000-th roots of unity. But brute-force computation of the $c_m$ is the kind of fool's errand that a human would never embark on, and that software stupidly attempts and fails to accomplish. (Maxima, Yacas, and Wolfram Alpha all try and fail.)
This is not homework.
 A: For $|x|<1$ we have $1/(x^n - 1) = - \sum_{k=0}^\infty x^{nk}$, so an antiderivative of this is
$ - \sum_{k=0}^\infty \frac{x^{nk+1}}{nk+1}$.  This can be written as a hypergeometric function:
$- x \ {}_2F_1\left(\frac{1}{n},1; 1+\frac{1}{n}; x^n\right)$.  
A: You can use the fact that, in a partial fraction decomposition, for a simple root $\alpha$ of the denominator (say $F = {P \over Q}$ where $(P,Q) = 1$) then the coefficient of ${1 \over X-\alpha}$ is ${P(\alpha) \over Q'(\alpha)}$. Since $X^{1000} - 1$ only has simple roots (the 1000th powers of unity), which you can express easily as $\omega^k, k \in \{0,\dots,999\}$ where $\omega = e^{2i\pi \over 1000}$. Then it's just a matter of computing a sum, since the integral of ${1 \over x-\alpha}$ is easy enough to compute.
Beware though, $\alpha$ is complex here, so the antiderivative is not just $\log(x-\alpha)$…. But another trick you can use is that you can naturally pair the roots of unity, as $\bar{\zeta} = \zeta^{-1}$ for $|\zeta| = 1$.
