Answer
$$\dfrac{123}{32}$$
Work Step by Step
The arc length can be calculated as follows:
$L=\int_{a}^{b} \sqrt {1+[f'(x)]^2} dx$
Re-arrange the given equation such that $[f'(x)]^2=\dfrac{(y-1)^2}{4y}$
After simplification, we have $L=\int_1^2 [1+\dfrac{(4y^6-1)^2}{16y^6}] dy $
Now, we will separate the terms and then integrate.
$$L=\int_1^2 y^3+\dfrac{y^{-3}}{4} dy \\=[ \dfrac{y^4}{4}-\dfrac{1}{8y^2} ]_1^{2} \\=\dfrac{123}{32}$$
