Answer
$${e^{\tan x}} + C$$
Work Step by Step
$$\eqalign{
& \int {{e^{\tan x}}{{\sec }^2}x} dx \cr
& {\text{substitute }}u = \tan x,{\text{ }}du = {\sec ^2}xdx \cr
& \int {{e^{\tan x}}{{\sec }^2}x} dx = \int {{e^u}du} \cr
& {\text{find the antiderivative}} \cr
& \int {{e^u}du} = {e^u} + C \cr
& {\text{write in terms of }}x,{\text{ replace }}u = \tan x \cr
& = {e^{\tan x}} + C \cr} $$