Answer
$$L = \frac{{17}}{{12}}$$
Work Step by Step
$$\eqalign{
& y = \frac{{{x^3}}}{6}{\text{ + }}\frac{1}{{2x}}{\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] \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) = \frac{{{x^3}}}{6}{\text{ + }}\frac{1}{{2x}}{\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^3}}}{6}{\text{ + }}\frac{1}{{2x}}} \right] \cr
& f'\left( x \right) = \frac{{3{x^2}}}{6} - \frac{1}{{2{x^2}}} \cr
& f'\left( x \right) = \frac{{{x^2}}}{2} - \frac{1}{{2{x^2}}} \cr
& {\text{Using the arc length formula}}{\text{, we have}} \cr
& L = \int_1^2 {\sqrt {1 + {{\left( {\frac{{{x^2}}}{2} - \frac{1}{{2{x^2}}}} \right)}^2}} } dx \cr
& {\text{simplifying}} \cr
& L = \int_1^2 {\sqrt {1 + {{\left( {\frac{{{x^2}}}{2}} \right)}^2} - \left( {\frac{{{x^2}}}{2}} \right)\left( {\frac{1}{{2{x^2}}}} \right) + {{\left( {\frac{1}{{2{x^2}}}} \right)}^2}} } dx \cr
& L = \int_1^2 {\sqrt {1 + {{\left( {\frac{{{x^2}}}{2}} \right)}^2} - \frac{1}{2} + {{\left( {\frac{1}{{2{x^2}}}} \right)}^2}} } dx \cr
& L = \int_1^2 {\sqrt {{{\left( {\frac{{{x^2}}}{2}} \right)}^2} + \frac{1}{2} + {{\left( {\frac{1}{{2{x^2}}}} \right)}^2}} } dx \cr
& {\text{factor}} \cr
& L = \int_1^2 {\sqrt {{{\left( {\frac{{{x^2}}}{2} + \frac{1}{{2{x^2}}}} \right)}^2}} } dx \cr
& L = \int_1^2 {\left( {\frac{{{x^2}}}{2} + \frac{1}{{2{x^2}}}} \right)} dx \cr
& L = \frac{1}{2}\int_1^2 {\left( {{x^2} + {x^{ - 2}}} \right)} dx \cr
& {\text{integrate}} \cr
& L = \frac{1}{2}\left( {\frac{{{x^3}}}{3} - \frac{1}{x}} \right)_1^2 \cr
& {\text{evaluate the limits}} \cr
& L = \frac{1}{2}\left( {\frac{{{{\left( 2 \right)}^3}}}{3} - \frac{1}{2}} \right) - \frac{1}{2}\left( {\frac{{{{\left( 1 \right)}^3}}}{3} - \frac{1}{1}} \right) \cr
& L = \frac{1}{2}\left( {\frac{{13}}{6}} \right) - \frac{1}{2}\left( { - \frac{2}{3}} \right) \cr
& L = \frac{{17}}{{12}} \cr} $$
