Answer
$$L = \ln \left( {\sqrt 2 + 1} \right)$$
Work Step by Step
$$\eqalign{
& y = \ln \left( {\sec x} \right),{\text{ }}0 \leqslant x \leqslant \frac{\pi }{4} \cr
& {\text{Differentiate}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\ln \left( {\sec x} \right)} \right] \cr
& \frac{{dy}}{{dx}} = \frac{{\sec x\tan x}}{{\sec x}} \cr
& \frac{{dy}}{{dx}} = \tan x \cr
& {\text{Use the Arc Length Formula}} \cr
& L = \int_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} } dx \cr
& L = \int_0^{\pi /4} {\sqrt {1 + {{\left( {\tan x} \right)}^2}} } dx \cr
& L = \int_0^{\pi /4} {\sqrt {1 + {{\tan }^2}x} } dx \cr
& L = \int_0^{\pi /4} {\sqrt {{{\sec }^2}x} } dx \cr
& L = \int_0^{\pi /4} {\sec x} dx \cr
& {\text{Integrate}} \cr
& L = \left[ {\ln \left| {\sec x + \tan x} \right|} \right]_0^{\pi /4} \cr
& L = \ln \left| {\sec \left( {\frac{\pi }{4}} \right) + \tan \left( {\frac{\pi }{4}} \right)} \right| - \ln \left| {\sec \left( 0 \right) + \tan \left( 0 \right)} \right| \cr
& L = \ln \left| {\sqrt 2 + 1} \right| - \ln \left| {1 + 0} \right| \cr
& L = \ln \left( {\sqrt 2 + 1} \right) \cr} $$
