Answer
$${2\sqrt 3 } - \frac{\pi }{6} - 2$$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /6} {{{\left( {\sec x + \tan x} \right)}^2}dx} \cr
& {\text{Expand the integrand}} \cr
& = \int_0^{\pi /6} {\left( {{{\sec }^2}x + 2\sec x\tan x + {{\tan }^2}x} \right)dx} \cr
& {\text{Use the identity ta}}{{\text{n}}^2}x + 1 = {\sec ^2}x \cr
& = \int_0^{\pi /6} {\left( {{{\sec }^2}x + 2\sec x\tan x + {{\sec }^2}x - 1} \right)dx} \cr
& = \int_0^{\pi /6} {\left( {2{{\sec }^2}x + 2\sec x\tan x - 1} \right)dx} \cr
& {\text{Integrate and evaluate}} \cr
& = \left( {2\tan x + 2\sec x - x} \right)_0^{\pi /6} \cr
& = \left( {2\tan \left( {\frac{\pi }{6}} \right) + 2\sec \left( {\frac{\pi }{6}} \right) - \frac{\pi }{6}} \right) - \left( {2\tan \left( 0 \right) + 2\sec \left( 0 \right) - 0} \right) \cr
& {\text{Simplifying gives}} \cr
& = 2\left( {\frac{{\sqrt 3 }}{3}} \right) + 2\left( {\frac{{2\sqrt 3 }}{3}} \right) - \frac{\pi }{6} - 0 - 2 + 0 \cr
& = {2\sqrt 3 }- \frac{\pi }{6} - 2 \cr} $$
