Find the length of the curve

$\displaystyle x = \frac{y^3}{3} + \frac{1}{4y}$ from $\displaystyle y=1$ to $\displaystyle y=3$

$\displaystyle g'(y) = y^2 + \frac{y^{-2}}{4}$

$\displaystyle L = \int \sqrt{1 + \left(y^2 + \frac{y^{-2}}{4}\right)^2}dy$

$\displaystyle L = \int \sqrt{1 + y^4 + \frac{1}{2} + \frac{y^{-4}}{16}}dy$

$\displaystyle L = \int \sqrt{1 + y^4 + \frac{3}{2} + \frac{y^{-4}}{16}}dy$

It says it in the book that 1 + \left(\frac{dx}{dy}\right)^2 is a perfect square but I'm not seeing that. Any help?

Edit: I think I found my error. :|