Sometimes it pays to persevere, or don't surrender early.

6xy = y^4 +3

Find the length of the curve from y=1 to y=4

So we use dy, because the boundaries are given in values of y.

Length of curve, L.

dL = sqrt[1 +(dx/dy)^2]dy

So we find dx/dy.

x = (y^4 +3)/(6y)

x = (y^3)/6 +1/(2y)

dx/dy = (y^2)/2 -1/(2y^2)

dx/dy = (y^4 -1) /(2y^2) ---------------***

So,

dL = sqrt[1 +[(y^4 -1)/(2y^2)]^2]dy

Now the perseverance. We expand the (dx/dy)^2, then continue from there.

dL = sqrt[1 +[(y^8 -2y^4 +1) /(4y^4)]dy

dL = sqrt[{(4y^4) +(y^8 -2y^4 +1)} / (4y^4)]dy

dL = sqrt[(y^8 +2y^4 +1) /(4y^4)]dy

dL = sqrt[(y^4 +1)^2 /(4y^4)]dy

dL = [(y^4 +1) /(2y^2)]dy

dL = [(y^2)/2 +1/(2y^2)]dy

dL = [(1/2)y^2 +(1/2)y^(-2)]dy

dL = (1/2)[y^2 +y^(-2)]dy

Now that can be integrated.

So,

L = (1/2)INT.(1-->4)[y^2 +y^(-2)]dy

L = (1/2)[(y^3)/3 -1/y]|(1-->4)

L = (1/2)[{(4^3)/3 -1/4} -{(1^3)/3 -1/1}]

L = (1/2)[64/3 -1/4 -1/3 +1]

L = 261/24 = 87/8 = 10.875 units long. --------------answer.