Answer
$$
\text { The length }= \int_{0}^{2\pi} (2-\cos \theta)^{-2}\sqrt{5-4\cos \theta} d \theta
$$
Work Step by Step
Since $r=f(\theta)=(2-\cos \theta)^{-1}$, then $f'(\theta)=-(2-\cos \theta)^{-2}\sin\theta$
The length is given by
$$
\text { The length }=\int_{0}^{2\pi} \sqrt{f(\theta)^{2}+f^{\prime}(\theta)^{2}} d \theta\\
=\int_{0}^{2\pi} \sqrt{(2-\cos \theta)^{-2}+(2-\cos \theta)^{-4}\sin^2\theta} d \theta\\
=\int_{0}^{2\pi} (2-\cos \theta)^{-2}\sqrt{(2-\cos \theta)^{2}+\sin^2\theta} d \theta\\
=\int_{0}^{2\pi} (2-\cos \theta)^{-2}\sqrt{5-4\cos \theta} d \theta
$$
