Answer
$L=16$
Work Step by Step
Given
$$
r=2(1+\cos \theta)
$$
The length is given by
\begin{align*}
L&=\int_{a}^{b} \sqrt{r^{2}+\left(\frac{d r}{ d \theta}\right)^{2}} d \theta\\
&= \int_{0}^{2 \pi} \sqrt{[2(1+\cos \theta)]^{2}+(-2 \sin \theta)^{2}} d \theta\\
&=\int_{0}^{2 \pi} \sqrt{4+8 \cos \theta+4 \cos ^{2} \theta+4 \sin ^{2} \theta} d \theta\\
&=\int_{0}^{2 \pi} \sqrt{8+8 \cos \theta} d \theta\\
&=\sqrt{8} \int_{0}^{2 \pi} \sqrt{1+\cos \theta} d \theta\\
&=\sqrt{8} \int_{0}^{2 \pi} \sqrt{2 \cos ^{2} \frac{\theta}{2}} d \theta\\
&=4\int_{0}^{2 \pi}\left|\cos \frac{\theta}{2}\right| d \theta\\
&=8\int_{0}^{\pi} \cos \frac{\theta}{2} d \theta\\
&=16 \sin \frac{\theta}{2}\bigg|_{0}^{\pi}\\
&=16
\end{align*}
