Chapter 5 - Section 5.4 - The Fundamental Theorem of Calculus - Exercises - Page 320: 16

$2\sqrt{3}-2-\frac{\pi}{6}$

\begin{aligned} \int_{0}^{\pi / 6}(\sec x+\tan x)^{2} d x&=\int_{0}^{\pi / 6}(\sec ^2x+2\sec(x)\tan(x)+\tan^2 x) d x\\ &= \int_{0}^{\pi / 6}(\sec ^2x+2\sec(x)\tan(x)+\sec^2 x-1) d x\\ &= \int_{0}^{\pi / 6}(2\sec ^2x+2\sec(x)\tan(x)-1) d x\\ &= 2\tan (x)+2\sec(x)-x\bigg|_{0}^{\pi/6}\\ &= \frac{2\sqrt{3}}{3} +\frac{4\sqrt{3}}{3}-2-\frac{\pi}{6}\\ &=2\sqrt{3}-2-\frac{\pi}{6}\end{aligned}