Answer
$$L = \ln \left( {\sqrt 2 + 1} \right)$$
Work Step by Step
$$\eqalign{
& y = \ln \left( {\sec x} \right),{\text{ for the interval }}0 \leqslant x \leqslant \pi /4 \cr
& \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\ln \left( {\sec x} \right)} \right] \cr
& \frac{{dy}}{{dx}} = \frac{{\sec x\tan x}}{{\sec x}} \cr
& \frac{{dy}}{{dx}} = \tan x \cr
& \cr
& {\text{Use the arc length formula }}\cr
& L = \int_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} } dx \cr
& \,\,\,{\text{We have the interval }}0 \leqslant x \leqslant \pi /4,\cr
& {\text{ then }}a = 0{\text{ and b}} = \pi /4 \cr
& L = \int_0^{\pi /4} {\sqrt {1 + {{\left( {\tan x} \right)}^2}} } dx \cr
& {\text{Simplify}} \cr
& L = \int_0^{\pi /4} {\sqrt {1 + {{\tan }^2}x} } dx \cr
& L = \int_0^{\pi /4} {\sqrt {{{\sec }^2}x} } dx \cr
& L = \int_0^{\pi /4} {\sec x} dx \cr
& {\text{Integrate}} \cr
& L = \left( {\ln \left| {\sec x + \tan x} \right|} \right)_0^{\pi /4} \cr
& L = \left( {\ln \left| {\sec \left( {\frac{\pi }{4}} \right) + \tan \left( {\frac{\pi }{4}} \right)} \right|} \right) - \left( {\ln \left| {\sec \left( 0 \right) + \tan \left( 0 \right)} \right|} \right) \cr
& L = \left( {\ln \left| {\sqrt 2 + 1} \right|} \right) - \left( {\ln \left| {1 + 0} \right|} \right) \cr
& L = \ln \left( {\sqrt 2 + 1} \right) \cr} $$
