Answer
$$L = \frac{{65}}{{32}}$$
Work Step by Step
$$\eqalign{
& x = 2{e^{\sqrt 2 y}} + \frac{1}{{16}}{e^{ - \sqrt 2 y}},{\text{ for }}0 \leqslant y \leqslant \frac{{\ln 2}}{{\sqrt 2 }} \cr
& {\text{Calculate the Arc Length }} \cr
& \frac{{dx}}{{dy}} = \frac{d}{{dy}}\left[ {2{e^{\sqrt 2 y}} + \frac{1}{{16}}{e^{ - \sqrt 2 y}}} \right] \cr
& \frac{{dx}}{{dy}} = 2\sqrt 2 {e^{\sqrt 2 y}} - \frac{{\sqrt 2 }}{{16}}{e^{ - \sqrt 2 y}} \cr
& {\text{Use the formula }}L = \int_c^d {\sqrt {1 + {{\left( {\frac{{dx}}{{dy}}} \right)}^2}} dy} \cr
& L = \int_0^{\frac{{\ln 2}}{{\sqrt 2 }}} {\sqrt {1 + {{\left( {2\sqrt 2 {e^{\sqrt 2 y}} - \frac{{\sqrt 2 }}{{16}}{e^{ - \sqrt 2 y}}} \right)}^2}} dy} \cr
& L = \int_0^{\frac{{\ln 2}}{{\sqrt 2 }}} {\sqrt {1 + {{\left( {2\sqrt 2 {e^{\sqrt 2 y}} - \frac{{\sqrt 2 }}{{16}}{e^{ - \sqrt 2 y}}} \right)}^2}} dy} \cr
& L = \int_0^{\frac{{\ln 2}}{{\sqrt 2 }}} {\sqrt {1 + 8{e^{2\sqrt 2 y}} - \frac{1}{2} + \frac{1}{{128}}{e^{ - 2\sqrt 2 y}}} dy} \cr
& L = \int_0^{\frac{{\ln 2}}{{\sqrt 2 }}} {\sqrt {8{e^{2\sqrt 2 y}} + \frac{1}{2} + \frac{1}{{128}}{e^{ - 2\sqrt 2 y}}} dy} \cr
& L = \int_0^{\frac{{\ln 2}}{{\sqrt 2 }}} {\sqrt {8\left( {{e^{2\sqrt 2 y}} + \frac{1}{{16}} + \frac{1}{{1024}}{e^{ - 2\sqrt 2 y}}} \right)} dy} \cr
& L = \int_0^{\frac{{\ln 2}}{{\sqrt 2 }}} {\sqrt {8{{\left( {{e^{\sqrt 2 y}} + \frac{1}{{32}}{e^{ - \sqrt 2 y}}} \right)}^2}} dy} \cr
& L = 2\sqrt 2 \int_0^{\frac{{\ln 2}}{{\sqrt 2 }}} {\left( {{e^{\sqrt 2 y}} + \frac{1}{{32}}{e^{ - \sqrt 2 y}}} \right)dy} \cr
& {\text{Integrating}} \cr
& L = 2\sqrt 2 \left[ {\frac{1}{{\sqrt 2 }}{e^{\sqrt 2 y}} - \frac{1}{{32\sqrt 2 }}{e^{ - \sqrt 2 y}}} \right]_0^{\frac{{\ln 2}}{{\sqrt 2 }}} \cr
& L = 2\sqrt 2 \left[ {\frac{1}{{\sqrt 2 }}{e^{\sqrt 2 \left( {\frac{{\ln 2}}{{\sqrt 2 }}} \right)}} - \frac{1}{{32\sqrt 2 }}{e^{ - \sqrt 2 \left( {\frac{{\ln 2}}{{\sqrt 2 }}} \right)}}} \right] - 2\sqrt 2 \left[ {\frac{1}{{\sqrt 2 }} - \frac{1}{{32\sqrt 2 }}} \right] \cr
& L = 2\sqrt 2 \left[ {\frac{1}{{\sqrt 2 }}\left( 2 \right) - \frac{1}{{32\sqrt 2 }}\left( {\frac{1}{2}} \right)} \right] - 2\sqrt 2 \left[ {\frac{1}{{\sqrt 2 }} - \frac{1}{{32\sqrt 2 }}} \right] \cr
& L = 2\left( {2 - \frac{1}{{64}}} \right) - 2\left( {1 - \frac{1}{{32}}} \right) \cr
& L = \frac{{65}}{{32}} \cr} $$
