Answer
$$L = \frac{{595}}{{144}}$$
Work Step by Step
$$\eqalign{
& y = \left( {{x^6} + 8} \right){\text{/}}\left( {16{x^2}} \right){\text{ from }}x = 2{\text{ to }}x = 3 \cr
& y = \frac{{{x^4}}}{{16}} + \frac{1}{{2{x^2}}} \cr
& {\text{Calculate }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\frac{{{x^4}}}{{16}} + \frac{1}{{2{x^2}}}} \right] \cr
& \frac{{dy}}{{dx}} = \frac{1}{4}{x^3} - \frac{1}{{{x^3}}} \cr
& {\text{Use }}L = \int_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} dx} \cr
& L = \int_2^3 {\sqrt {1 + {{\left( {\frac{1}{4}{x^3} - \frac{1}{{{x^3}}}} \right)}^2}} dx} \cr
& L = \int_2^3 {\sqrt {1 + \frac{1}{{16}}{x^6} - 2\left( {\frac{1}{4}{x^3}} \right)\left( {\frac{1}{{{x^3}}}} \right) + \frac{1}{{{x^6}}}} dx} \cr
& L = \int_2^3 {\sqrt {\frac{1}{{16}}{x^6} + \frac{1}{2} + \frac{1}{{{x^6}}}} dx} \cr
& L = \int_2^3 {\sqrt {{{\left( {\frac{1}{4}{x^3} + \frac{1}{{{x^3}}}} \right)}^2}} dx} \cr
& L = \int_2^3 {\left( {\frac{1}{4}{x^3} + \frac{1}{{{x^3}}}} \right)dx} \cr
& {\text{Integrating}} \cr
& L = \left[ {\frac{1}{{16}}{x^4} - \frac{1}{{2{x^2}}}} \right]_2^3 \cr
& L = \left[ {\frac{1}{{16}}{{\left( 3 \right)}^4} - \frac{1}{{2{{\left( 3 \right)}^2}}}} \right] - \left[ {\frac{1}{{16}}{{\left( 2 \right)}^4} - \frac{1}{{2{{\left( 2 \right)}^2}}}} \right] \cr
& L = \frac{{721}}{{144}} - \frac{7}{8} \cr
& L = \frac{{595}}{{144}} \cr} $$
