Answer
$\int \frac{sec~\theta~tan~\theta}{1+sec~\theta}~d\theta = ln \vert 1+sec~\theta \vert+C$
Work Step by Step
$\int \frac{sec~\theta~tan~\theta}{1+sec~\theta}~d\theta$
Let $u = sec~\theta$
$\frac{du}{d\theta} = sec~\theta~tan~\theta$
$d\theta = \frac{du}{sec~\theta~tan~\theta}$
$\int~\frac{sec~\theta~tan~\theta}{1+u}\frac{du}{sec~\theta~tan~\theta}$
$=\int~\frac{1}{1+u}~du$
$=ln \vert 1+u \vert+C$
$=ln \vert 1+sec~\theta \vert+C$
