Answer
$$\tan \left( {\ln \theta } \right) + \frac{{{{\tan }^3}\left( {\ln \theta } \right)}}{3} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{{\sec }^4}\ln \left( \theta \right)}}{\theta }d\theta } \cr
& {\text{substitute }}u = \ln \theta ,{\text{ }}du = \frac{1}{\theta }d\theta \cr
& = \int {{{\sec }^4}u} du \cr
& {\text{split off }}{\sec ^4}u \cr
& = \int {{{\sec }^2}u} {\sec ^2}udu \cr
& {\text{pythagorean identity}} \cr
& = \int {\left( {1 + {{\tan }^2}u} \right)} {\sec ^2}udu \cr
& = \int {\left( {{{\sec }^2}u + {{\tan }^2}u{{\sec }^2}u} \right)} du \cr
& {\text{evaluate}} \cr
& = \tan u + \frac{{{{\tan }^3}u}}{3} + C \cr
& {\text{replace }}u = \ln \theta \cr
& = \tan \left( {\ln \theta } \right) + \frac{{{{\tan }^3}\left( {\ln \theta } \right)}}{3} + C \cr} $$
