Answer
$$L = \frac{{123}}{{32}}$$
Work Step by Step
$$\eqalign{
& x = \frac{{{y^4}}}{4} + \frac{1}{{8{y^2}}},{\text{ for }}1 \leqslant y \leqslant 2 \cr
& {\text{Calculate the Arc Length }} \cr
& \frac{{dx}}{{dy}} = \frac{d}{{dy}}\left[ {\frac{{{y^4}}}{4} + \frac{1}{{8{y^2}}}} \right] \cr
& \frac{{dx}}{{dy}} = \frac{{4{y^3}}}{4} - \frac{{2{y^{ - 3}}}}{8} \cr
& \frac{{dx}}{{dy}} = {y^3} - \frac{1}{4}{y^{ - 3}} \cr
& {\text{Use the formula }}L = \int_c^d {\sqrt {1 + {{\left( {\frac{{dx}}{{dy}}} \right)}^2}} dy} \cr
& L = \int_1^2 {\sqrt {1 + {{\left( {{y^3} - \frac{1}{4}{y^{ - 3}}} \right)}^2}} dy} \cr
& L = \int_1^2 {\sqrt {1 + {y^6} - \frac{1}{2}{y^3}{y^{ - 3}} + \frac{1}{{16}}{y^{ - 6}}} dy} \cr
& L = \int_1^2 {\sqrt {{y^6} + \frac{1}{2} + \frac{1}{{16}}{y^{ - 6}}} dy} \cr
& L = \int_1^2 {\sqrt {{{\left( {{y^3} + \frac{1}{4}{y^{ - 3}}} \right)}^2}} dy} \cr
& L = \int_1^2 {\left( {{y^3} + \frac{1}{4}{y^{ - 3}}} \right)dy} \cr
& {\text{Integrating}} \cr
& L = \left[ {\frac{1}{4}{y^4} - \frac{1}{8}{y^{ - 2}}} \right]_1^2 \cr
& L = \left[ {\frac{1}{4}{{\left( 2 \right)}^4} - \frac{1}{8}{{\left( 2 \right)}^{ - 2}}} \right] - \left[ {\frac{1}{4}{{\left( 1 \right)}^4} - \frac{1}{8}{{\left( 1 \right)}^{ - 2}}} \right] \cr
& L = \frac{{127}}{{32}} - \frac{1}{8} \cr
& L = \frac{{123}}{{32}} \cr} $$
