Answer
$$L = \frac{{2055}}{{64}}$$
Work Step by Step
$$\eqalign{
& x = \frac{1}{8}{y^4} + \frac{1}{4}{y^{ - 2}}{\text{ from }}y = 1{\text{ to }}y = 4 \cr
& {\text{Calculate }}\frac{{dx}}{{dy}} \cr
& \frac{{dx}}{{dy}} = \frac{d}{{dy}}\left[ {\frac{1}{8}{y^4} + \frac{1}{4}{y^{ - 2}}} \right] \cr
& \frac{{dy}}{{dx}} = \frac{4}{8}{y^3} - \frac{2}{4}{y^{ - 3}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{2}{y^3} - \frac{1}{2}{y^{ - 3}} \cr
& {\text{Use }}L = \int_a^b {\sqrt {1 + {{\left( {\frac{{dx}}{{dy}}} \right)}^2}} dy} \cr
& L = \int_1^4 {\sqrt {1 + {{\left( {\frac{1}{2}{y^3} - \frac{1}{2}{y^{ - 3}}} \right)}^2}} dy} \cr
& L = \int_1^4 {\sqrt {1 + {{\left( {\frac{1}{2}{y^3}} \right)}^2} - 2\left( {\frac{1}{2}{y^3}} \right)\left( {\frac{1}{2}{y^{ - 3}}} \right) + {{\left( {\frac{1}{2}{y^{ - 3}}} \right)}^2}} dy} \cr
& L = \int_1^4 {\sqrt {{{\left( {\frac{1}{2}{y^3}} \right)}^2} + \frac{1}{2} + {{\left( {\frac{1}{2}{y^{ - 3}}} \right)}^2}} dy} \cr
& L = \int_1^4 {\sqrt {{{\left( {\frac{1}{2}{y^3} + \frac{1}{2}{y^{ - 3}}} \right)}^2}} dy} \cr
& L = \int_1^4 {\left( {\frac{1}{2}{y^3} + \frac{1}{2}{y^{ - 3}}} \right)dy} \cr
& {\text{Integrating}} \cr
& L = \left[ {\frac{1}{8}{y^4} - \frac{1}{4}{y^{ - 2}}} \right]_1^4 \cr
& L = \left[ {\frac{1}{8}{{\left( 4 \right)}^4} - \frac{1}{4}{{\left( 4 \right)}^{ - 2}}} \right] - \left[ {\frac{1}{8}{{\left( 1 \right)}^4} - \frac{1}{4}{{\left( 1 \right)}^{ - 2}}} \right] \cr
& L = \frac{{2047}}{{64}} + \frac{1}{8} \cr
& L = \frac{{2055}}{{64}} \cr} $$
