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

$\dfrac{53}{6}$
Formula to calculate the arc length is: $l=\int_p^q \sqrt {1+[f'(x)]^2} dy$ The given equation can be re-written as: $f'(x)=y^2-\dfrac{1}{4y^2}$ and $(f'(x))^2=y^4+\dfrac{1}{16 y^4}-\dfrac{1}{2}$ Also, $1+(f'(x))^2=[y^2+\dfrac{1}{4y^2}]^2 or, \sqrt {1+(f'(x))^2}=y^2+\dfrac{1}{4y^2}$ Since, we know $\int x^n dx=\dfrac{x^{n+1}}{n+1}+C$ Now, $l=\int_1^3 [y^2+\dfrac{1}{4y^2}] dy =[\dfrac{y^3}{3}-\dfrac{1}{4y}]_1^3=\dfrac{53}{6}$