## Thomas' Calculus 13th Edition

$\dfrac{13}{4}$
The formula to calculate the arc length is as follows: $L=\int_m^n \sqrt {1+[f'(x)]^2} dx$ Re-write the equation as follows: $[f'(x)]^2=\dfrac{(y-1)^2}{4y}$ This implies that $L=\int_{2}^{3} [1+\dfrac{(y^4-1)^2}{4y^4}] dy$ Separate the terms and integrate as follows: $L=\int_{2}^{3} (y^2/2)+\dfrac{y^{-2}}{2} dy \\=[ \dfrac{y^3}{6}-\dfrac{1}{2y} ]_2^{3}=\dfrac{13}{4}$