Answer
$$\frac{1}{2}\sec {e^x}\tan {e^x} + \frac{1}{2}\ln \left| {\sec {e^x} + \tan {e^x}} \right| + C $$
Work Step by Step
$$\eqalign{
& \int {{e^x}{{\sec }^3}{e^x}} dx \cr
& {\text{Let }}u = {e^x},\,\,\,\,du = {e^x}dx,\,\,\,\,dx = \frac{{du}}{{{e^x}}} \cr
& {\text{Then}}{\text{,}} \cr
& \int {{e^x}{{\sec }^3}{e^x}} dx = \int {{e^x}{{\sec }^3}u} \left( {\frac{{du}}{{{e^x}}}} \right) \cr
& {\text{Cancel common factor }}{e^x} \cr
& = \int {{{\sec }^3}u} du \cr
& \cr
& {\text{From example 6, we obtain }} \cr
& \int {{{\sec }^3}xdx} = \frac{1}{2}\sec x\tan x + \frac{1}{2}\ln \left| {\sec x + \tan x} \right| + C \cr
& {\text{Then}}{\text{,}} \cr
& \cr
& = \frac{1}{2}\sec u\tan u + \frac{1}{2}\ln \left| {\sec u + \tan u} \right| + C \cr
& {\text{write in terms of }}x,{\text{ use }}u = {e^x} \cr
& = \frac{1}{2}\sec {e^x}\tan {e^x} + \frac{1}{2}\ln \left| {\sec {e^x} + \tan {e^x}} \right| + C \cr} $$
