Answer
$$L = \frac{{123}}{{32}}$$
Work Step by Step
$$\eqalign{
& y = \frac{{{x^4}}}{4} + \frac{1}{{8{x^2}}}{\text{ on }}\left[ {1,2} \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{{{x^4}}}{4} + \frac{1}{{8{x^2}}}{\text{ and }}\left[ {1,2} \right] \to a = 1{\text{ and }}b = 2.{\text{ then}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{{x^4}}}{4} + \frac{1}{{8{x^2}}}} \right] \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{{{x^4}}}{4} + \frac{1}{8}{x^{ - 2}}} \right] \cr
& f'\left( x \right) = \frac{{4{x^3}}}{4} - \frac{2}{8}{x^{ - 3}} \cr
& f'\left( x \right) = {x^3} - \frac{1}{4}{x^{ - 3}} \cr
& {\text{Using the arc length formula}}{\text{, we have}} \cr
& L = \int_1^2 {\sqrt {1 + {{\left( {{x^3} - \frac{1}{4}{x^{ - 3}}} \right)}^2}} } dx \cr
& {\text{simplifying}} \cr
& L = \int_1^2 {\sqrt {1 + {{\left( {{x^3}} \right)}^2} - 2\left( {{x^3}} \right)\left( {\frac{1}{4}{x^{ - 3}}} \right) + {{\left( {\frac{1}{4}{x^{ - 3}}} \right)}^2}} } dx \cr
& L = \int_1^2 {\sqrt {1 + {{\left( {{x^3}} \right)}^2} - \frac{1}{2} + {{\left( {\frac{1}{4}{x^{ - 3}}} \right)}^2}} } dx \cr
& L = \int_1^2 {\sqrt {{{\left( {{x^3}} \right)}^2} + \frac{1}{2} + {{\left( {\frac{1}{4}{x^{ - 3}}} \right)}^2}} } dx \cr
& {\text{factoring }} \cr
& L = \int_1^2 {\sqrt {{{\left( {{x^3} + \frac{1}{4}{x^{ - 3}}} \right)}^2}} } dx \cr
& L = \int_1^2 {\left( {{x^3} + \frac{1}{4}{x^{ - 3}}} \right)} dx \cr
& {\text{integrate and evaluate}} \cr
& L = \left( {\frac{{{x^4}}}{4} - \frac{{{x^{ - 2}}}}{8}} \right)_1^2 \cr
& L = \left( {\frac{{{{\left( 2 \right)}^4}}}{4} - \frac{{{{\left( 2 \right)}^{ - 2}}}}{8}} \right) - \left( {\frac{{{{\left( 1 \right)}^4}}}{4} - \frac{{{{\left( 1 \right)}^{ - 2}}}}{8}} \right) \cr
& L = \frac{{127}}{{32}} - \frac{1}{8} \cr
& L = \frac{{123}}{{32}} \cr} $$
