## University Calculus: Early Transcendentals (3rd Edition)

$$\dfrac{13}{4}$$
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 $L=\int_{2}^{3} [1+\dfrac{(y^4-1)^2}{4y^4}] \space dy$ Now, we will separate the terms and then integrate: $$L=\int_{2}^{3} (\dfrac{y^2}{2})+\dfrac{y^{-(2)}}{2} dy \\=[ \dfrac{y^3}{6}-\dfrac{1}{2y} ]_2^{3} \\=\dfrac{13}{4}$$