## Thomas' Calculus 13th Edition

$\dfrac{53}{6}$
We know that the formula to calculate the arc length is defined as: $L=\int_m^n \sqrt {1+[f'(x)]^2} dx$ Re-write the equation as follows: $f'(x)=(y^2-\dfrac{1}{4y^2})$ and $(f'(x))^2=y^4+\dfrac{1}{16 y^4}-\dfrac{1}{2}$ This implies that $L=1+(f'(x))^2=[y^2+\dfrac{1}{4y^2}]^2 \\ \implies L=\sqrt {1+(f'(x))^2}=y^2+\dfrac{1}{4y^2}$ Then, $L=\int_1^3 [y^2+\dfrac{1}{4y^2}] dy$ and $[\dfrac{y^{2+1}}{(2+1)}-\dfrac{1}{4y}]_1^3=[\dfrac{y^{3}}{3}-\dfrac{1}{4y}]_1^3=\dfrac{53}{6}$