Answer
$\dfrac{13}{12} $
Work Step by Step
Since, $L=\int_{1}^{2}\sqrt{1+(\dfrac{dy}{dx})^2}dx$
Thus, $L=\int_{1}^{2} \sqrt{1+\dfrac{1}{16}(y^{4}-\dfrac{1}{2}+\dfrac{1}{y^4})} dy$
or, $L=(\dfrac{1}{2}y^3-y^{-1})_1^{2}$
Thus, $L =\dfrac{13}{12} $
