y' = x^{2} - \dfrac{1}{4x^{2}}. \displaystyle 1 + (y')^{2} = 1 + \left(x^{2} - \dfrac{1}{4x^{2}}\right)^{2} = 1 + x^{4} - \dfrac{1}{2} + \dfrac{1}{16x^{4}} = \left(x^{2} + \dfrac{1}{4x^{2}}\right)^{2}

\displaystyle L = \int_{1}^{2} \left(x^{2} + \dfrac{1}{4x^{2}}\right) dx = \left[\dfrac{x^{3}}{3} - \dfrac{1}{4x}\right]_{1}^{2}

\displaystyle L = \left(\dfrac{8}{3} - \dfrac{1}{8}\right) - \left(\dfrac{1}{3} - \dfrac{1}{4}\right) = \dfrac{7}{3} + \dfrac{1}{8} = \dfrac{56 + 3}{24} = \dfrac{59}{24}