Answer
$\frac{1}{4}\ln {|\sec4x +\tan4x|} +C$
Work Step by Step
Use u-substitution and a Basic integration formula.
$\int sec4xdx$
let $u=4x$, $du=4dx$
$\frac{1}{4}\int secu du$, Substitute u
$\frac{1}{4}\ln{|\sec u +\tan u|}+C$, Integrate
$\frac{1}{4}\ln{|\sec4x+ \tan4x|}+C$
