At New York’s Kennedy Airport today, an individual was arrested trying to board a flight while in possession of a ruler, a protractor, a setsquare, a slide rule, and a calculator. At a morning press conference, Attorney General John Ashcroft said he believes the man is a member of the notorious al-gebra movement. He is being charged by the FBI with carrying weapons of math instruction.
“Al-gebra is a fearsome cult,” Ashcroft said. “They desire average solutions by means and extremes, and sometimes go off on tangents in a search of absolute value. They use secret code names like ‘x’ and ‘y’ …

@Jeff: I have to leave for a while. Here is my answer.
Considering [this path](https://i.sstatic.net/mj8RA.png)
$$
\begin{align}
\int_\gamma\frac{\mathrm{d}z}{z^5+a^5}
&=\left(1-e^{2\pi i/5}\right)\int_0^\infty\frac{\mathrm{d}x}{x^5+a^5}\\
&=2\pi i\text{ Res}\left(\frac{1}{z^5+a^5},e^{\pi i/5}\right)\\
&=2\pi i\lim_{z\to ae^{\pi i/5}}\frac{z-ae^{\pi i/5}}{z^5+a^5}\\
&=2\pi i\lim_{z\to ae^{\pi i/5}}\frac{1}{5z^4}\\
&=\frac{2\pi i}{5a^4e^{4\pi i/5}}
\end{align}
$$
Therefore,
$$
\begin{align}
\int_0^\infty\frac{\mathrm{d}x}{x^5+a^5}