Answer
$$\frac{1}{4}\ln \left| {\sec 4x + \tan 4x} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\sec 4x} dx \cr
& {\text{substitute }}u = 4x,{\text{ }}du = 4dx \cr
& \frac{1}{4}du = dx \cr
& \int {\sec 4x} dx = \frac{1}{4}\int {\sec u} du \cr
& {\text{find the antiderivative }}\left( {{\text{use formula 22 page 503}}} \right) \cr
& = \frac{1}{4}\ln \left| {\sec u + \tan u} \right| + C \cr
& {\text{write in terms of }}x,{\text{ replace }}u = 4x \cr
& = \frac{1}{4}\ln \left| {\sec 4x + \tan 4x} \right| + C \cr} $$
