Answer
$$L = \frac{3}{2}$$
Work Step by Step
$$\eqalign{
& y = \frac{1}{2}\left( {{e^x} + {e^{ - x}}} \right){\text{ on }}\left[ { - \ln 2,ln2} \right] \cr
& {\text{Definition of Arc Length for }}y = f\left( x \right): \cr
& {\text{Let }}f{\text{ have a continuous first derivative on the interval }}\left[ {a,b} \right]{\text{ The length of the curve }} \cr
& {\text{from }}\left( {a,f\left( a \right)} \right){\text{ to }}\left( {b,f\left( b \right)} \right){\text{ is }}L = \int_a^b {\sqrt {1 + f'{{\left( x \right)}^2}} } dx \cr
& {\text{Notice that }}y = f\left( x \right) = \frac{1}{2}\left( {{e^x} + {e^{ - x}}} \right){\text{ and }}\left[ { - \ln 2,ln2} \right] \to a = - \ln 2{\text{ and }}b = \ln 2.{\text{ then}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{1}{2}\left( {{e^x} + {e^{ - x}}} \right)} \right] \cr
& f'\left( x \right) = \frac{1}{2}\left( {{e^x} - {e^{ - x}}} \right) \cr
& {\text{Using the arc length formula}}{\text{, we have}} \cr
& L = \int_{ - \ln 2}^{\ln 2} {\sqrt {1 + {{\left( {\frac{1}{2}\left( {{e^x} - {e^{ - x}}} \right)} \right)}^2}} } dx \cr
& {\text{simplifying}} \cr
& L = \int_{ - \ln 2}^{\ln 2} {\sqrt {1 + \frac{{{{\left( {{e^x} - {e^{ - x}}} \right)}^2}}}{4}} } dx \cr
& L = \int_{ - \ln 2}^{\ln 2} {\frac{1}{2}\sqrt {4 + {{\left( {{e^x} - {e^{ - x}}} \right)}^2}} } dx \cr
& L = \frac{1}{2}\int_{ - \ln 2}^{\ln 2} {\sqrt {4 + {e^{2x}} - 2{e^x}{e^{ - x}} + {e^{ - 2x}}} } dx \cr
& L = \frac{1}{2}\int_{ - \ln 2}^{\ln 2} {\sqrt {2 + {e^{2x}} + {e^{ - 2x}}} } dx \cr
& {\text{factoring }}2 + {e^{2x}} + {e^{ - 2x}} \cr
& L = \frac{1}{2}\int_{ - \ln 2}^{\ln 2} {\sqrt {{{\left( {{e^x} + {e^{ - x}}} \right)}^2}} } dx \cr
& L = \frac{1}{2}\int_{ - \ln 2}^{\ln 2} {\left( {{e^x} + {e^{ - x}}} \right)} dx \cr
& {\text{integrate and evaluate}} \cr
& L = \frac{1}{2}\left( {{e^x} - {e^{ - x}}} \right)_{ - \ln 2}^{\ln 2} \cr
& L = \frac{1}{2}\left( {{e^{\ln 2}} - {e^{ - \ln 2}}} \right) - \frac{1}{2}\left( {{e^{ - \ln 2}} - {e^{\ln 2}}} \right) \cr
& L = \frac{1}{2}\left( {2 - \frac{1}{2}} \right) - \frac{1}{2}\left( {\frac{1}{2} - 2} \right) \cr
& L = \frac{3}{4} + \frac{3}{4} \cr
& L = \frac{3}{2} \cr} $$
