#### Answer

$\dfrac{53}{6}$

#### Work Step by Step

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 $[\dfrac{y^{2+1}}{(2+1)}-\dfrac{1}{4y}]_1^3=[\dfrac{y^{3}}{3}-\dfrac{1}{4y}]_1^3=\dfrac{53}{6}$