Find the length of arc of a curve$\displaystyle 6xy = y^4 + 3$:

from the point where y = 1 to y = 4;

$\displaystyle

x = \frac{y^4 + 3}{6y}$

$\displaystyle \frac{dx}{dy} = \frac{y^2}{2} - \frac{3}{y^2} = \frac{y^4 - 6}{4y^4} $

the problem is that i can't integrate:

$\displaystyle \int \sqrt{1 + \frac{(y^4 - 6)^2}{4y^4}}dy$