Answer
$$s = \frac{{14}}{3}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{1}{6}{x^3} + \frac{1}{{2x}},{\text{ }}\left[ {1,3} \right] \cr
& {\text{Differentiate}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{1}{6}{x^3} + \frac{1}{{2x}}} \right] \cr
& f'\left( x \right) = \frac{{3{x^2}}}{6} - \frac{{{x^{ - 2}}}}{2} \cr
& f'\left( x \right) = \frac{1}{2}{x^2} - \frac{1}{{2{x^2}}} \cr
& {\text{Use the arc length formula}} \cr
& s = \int_a^b {\sqrt {1 + {{\left[ {f'\left( x \right)} \right]}^2}} dx} \cr
& s = \int_1^3 {\sqrt {1 + {{\left( {\frac{1}{2}{x^2} - \frac{1}{{2{x^2}}}} \right)}^2}} dx} \cr
& s = \int_1^3 {\sqrt {1 + \frac{{{x^4}}}{4} - \frac{1}{2} + \frac{1}{{4{x^4}}}} dx} \cr
& s = \int_1^3 {\sqrt {\frac{{{x^4}}}{4} + \frac{1}{2} + \frac{1}{{4{x^4}}}} dx} \cr
& {\text{Factoring}} \cr
& s = \int_1^3 {\sqrt {{{\left( {\frac{{{x^2}}}{2} + \frac{1}{{2{x^2}}}} \right)}^2}} dx} \cr
& s = \int_1^3 {\left( {\frac{{{x^2}}}{2} + \frac{1}{{2{x^2}}}} \right)dx} \cr
& {\text{Integrating}} \cr
& s = \left[ {\frac{{{x^3}}}{6} - \frac{1}{{2x}}} \right]_1^3 \cr
& s = \frac{{13}}{3} + \frac{1}{3} \cr
& s = \frac{{14}}{3} \cr} $$
