Answer
$ \dfrac{13}{12}$
Work Step by Step
Integrate the integral to calculate the arc length as follows:
$l= \int_{A}^{B} \sqrt {1+(\dfrac{dy}{dx})^2}$
or, $ =\int_{1}^{2}[\sqrt {1+\dfrac{y^4}{16}-\dfrac{1}{2} +\dfrac {1}{y^4}]} dy$
or, $=[\dfrac{y^3}{12}-\dfrac{1}{y}]_1^2$
or, $= \dfrac{13}{12}$
