Answer
$${t^3} + \frac{1}{2}\tan 2t + C$$
Work Step by Step
$$\eqalign{
& \int {\left( {3{t^2} + {{\sec }^2}2t} \right)} dt \cr
& {\text{sum rule}} \cr
& = \int {3{t^2}} dt + \int {{{\sec }^2}2t} dt \cr
& {\text{use the power rule and}} \cr
& {\text{use the formula for indefinite integrals of trigonometric functions}} \cr
& = 3\left( {\frac{{{t^3}}}{3}} \right) + \frac{1}{2}\tan 2t + C \cr
& {\text{simplify}} \cr
& = {t^3} + \frac{1}{2}\tan 2t + C \cr
& {\text{check by differentiation}} \cr
& {\text{ = }}\frac{d}{{dt}}\left( {{t^3} + \frac{1}{2}\tan 2t + C} \right) \cr
& {\text{ = }}\frac{d}{{dt}}\left( {{t^3}} \right) + \frac{d}{{dt}}\left( {\frac{1}{2}\tan 2t} \right) + \frac{d}{{dt}}\left( C \right) \cr
& = 3{t^2} + \frac{1}{2}\left( {{{\sec }^2}2t} \right)\left( 2 \right) + 0 \cr
& = 3{t^2} + {\sec ^2}2t \cr} $$
