Answer
$r=\ln|\tan t+1|+4$
When C=$4, $the graph passes through $(\pi, 4)$
Work Step by Step
$r=\displaystyle \int\frac{\sec^{2}t}{\tan t+1}dt$
$\left[\begin{array}{l}
u=\tan t+1\\
du=\sec^{2}tdt
\end{array}\right]$
$r=\displaystyle \int\frac{1}{u}du=\ln|u|+C$
$r=\ln|\tan t+1|+C$
$(\pi, 4) $is on the graph, so we find C
$4=\ln|0+1|+C$
$C=4$
$r=\ln|\tan t+1|+4$
