Answer
$ - \csc \theta + \cot \theta + \theta + C$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{\sec \theta + 1}}} d\theta \cr
& {\text{Multiplty the numerator and denominator by the conjugate of }} \cr
& {\text{the denominator}} \cr
& = \int {\frac{1}{{\sec \theta + 1}} \times \frac{{\sec \theta - 1}}{{\sec \theta - 1}}} d\theta \cr
& = \int {\frac{{\sec \theta - 1}}{{{{\sec }^2}\theta - 1}}} d\theta \cr
& {\text{Use the pythagorean identity }}1 + {\tan ^2}\theta = {\sec ^2}\theta \cr
& \int {\frac{{\sec \theta - 1}}{{{{\sec }^2}\theta - 1}}} d\theta = \int {\frac{{\sec \theta - 1}}{{{{\tan }^2}\theta }}} d\theta \cr
& {\text{Distribute the numerator}} \cr
& = \int {\left( {\frac{{\sec \theta }}{{{{\tan }^2}\theta }} - \frac{1}{{{{\tan }^2}\theta }}} \right)} d\theta \cr
& = \int {\left( {\cot \theta \csc \theta - {{\cot }^2}\theta } \right)} d\theta \cr
& {\text{Use the pythagorean identity }}{\cot ^2}\theta = {\csc ^2}\theta - 1 \cr
& = \int {\left( {\cot \theta \csc \theta - \left( {{{\csc }^2}\theta - 1} \right)} \right)} d\theta \cr
& = \int {\left( {\cot \theta \csc \theta - {{\csc }^2}\theta + 1} \right)} d\theta \cr
& {\text{Integrating}} \cr
& = - \csc \theta - \left( { - \cot \theta } \right) + \theta + C \cr
& = - \csc \theta + \cot \theta + \theta + C \cr} $$
