Answer
$\frac{\sqrt {1+\pi^2}}{\pi}-\frac{\sqrt {1+4\pi^2}}{2\pi}+ln[\frac{2\pi\sqrt {1+4\pi^2}}{\pi+\sqrt {1+\pi^2}}]$
Work Step by Step
$r=\frac{1}{\theta}$ ; $\pi\leq \theta \leq 2 \pi$
$r'=-\frac{1}{\theta^{2}}$ ; $\pi\leq \theta \leq 2 \pi$
Length of the curve is given as:
$L=\int_{\pi}^{2 \pi}\sqrt {(\frac{1}{\theta})^2+(-\frac{1}{\theta^{2}})^2} dt$
$=\int_{\pi}^{2 \pi}\sqrt {\frac{1}{\theta^2}+\frac{1}{\theta^4}} dt$
$=\frac{\sqrt {1+\pi^2}}{\pi}-\frac{\sqrt {1+4\pi^2}}{2\pi}+ln[\frac{2\pi\sqrt {1+4\pi^2}}{\pi+\sqrt {1+\pi^2}}]$
