## Calculus 10th Edition

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