Answer
$\dfrac{285}{8} $
Work Step by Step
Since, $L=\int_{1}^{32}\sqrt{1+(\dfrac{dy}{dx})^2}dx$
Thus, $L=\int_{1}^{32} \sqrt{1+\dfrac{1}{4}(x^{2/5}-2+x^{-2/5}} dx=\int_{1}^{32} (x^{1/5}+x^{-1/5})dx$
or, $L=(\dfrac{1}{2})[(5/6)x^{6/5}+(5/4)x^{4/5}]_1^{32}$
Thus, $L =\dfrac{285}{8} $
