Answer
$$L = 1$$
Work Step by Step
$$\eqalign{
& y = {\cosh ^{ - 1}}x{\text{ on }}\left[ {\sqrt 2 ,\sqrt 5 } \right] \cr
& {\text{Definition of Arc Length for }}y = f\left( x \right): \cr
& {\text{Let }}f{\text{ have a continuous first derivative on the interval }}\left[ {a,b} \right] \cr
& {\text{ The length of the curve from }}\left( {a,f\left( a \right)} \right){\text{ to }}\left( {b,f\left( b \right)} \right){\text{ is }}L = \int_a^b {\sqrt {1 + f'{{\left( x \right)}^2}} } dx \cr
& {\text{Notice that }}y = f\left( x \right) = {\cosh ^{ - 1}}x{\text{ and }}\left[ {\sqrt 2 ,\sqrt 5 } \right] \to a = \sqrt 2 {\text{ and }}b = \sqrt 5 .{\text{ then}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {{{\cosh }^{ - 1}}x} \right] \cr
& f'\left( x \right) = \frac{1}{{\sqrt {{x^2} - 1} }} \cr
& {\text{Using the arc length formula}}{\text{, we have}} \cr
& L = \int_{\sqrt 2 }^{\sqrt 5 } {\sqrt {1 + {{\left( {\frac{1}{{\sqrt {{x^2} - 1} }}} \right)}^2}} } dx \cr
& {\text{simplifying}} \cr
& L = \int_{\sqrt 2 }^{\sqrt 5 } {\sqrt {1 + \frac{1}{{{x^2} - 1}}} } dx \cr
& L = \int_{\sqrt 2 }^{\sqrt 5 } {\sqrt {\frac{{{x^2} - 1 + 1}}{{{x^2} - 1}}} } dx \cr
& L = \int_{\sqrt 2 }^{\sqrt 5 } {\sqrt {\frac{{{x^2}}}{{{x^2} - 1}}} } dx \cr
& L = \int_{\sqrt 2 }^{\sqrt 5 } {\frac{x}{{\sqrt {{x^2} - 1} }}} dx \cr
& L = \frac{1}{2}\int_{\sqrt 2 }^{\sqrt 5 } {{{\left( {{x^2} - 1} \right)}^{ - 1/2}}\left( {2x} \right)} dx \cr
& {\text{integrate}} \cr
& L = \frac{1}{2}\left( {\frac{{{{\left( {{x^2} - 1} \right)}^{1/2}}}}{{1/2}}} \right)_{\sqrt 2 }^{\sqrt 5 } \cr
& L = \left( {\sqrt {{x^2} - 1} } \right)_{\sqrt 2 }^{\sqrt 5 } \cr
& {\text{evaluate the limits}} \cr
& L = \sqrt {{{\left( {\sqrt 5 } \right)}^2} - 1} - \sqrt {{{\left( {\sqrt 2 } \right)}^2} - 1} \cr
& L = \sqrt 4 - \sqrt 1 \cr
& L = 1 \cr} $$
