Finding Arc Length of x = (y^4)/8 + 1/(4y^2) from 1 to 2

[MHrtz]

Homework Statement

Determine the arc length of the function on the given interval

x = (y^4)/8 + 1/(4y^2) from 1 to 2

The arc length formula

\int (f'(x)² + 1)^.5 dx

The Attempt at a Solution

I used the arc length formula but don't know where to go from here. Usually these problems can't be done unless a form of cancellation takes place and I can't seem to find it. Below is what I input in the formula but I can not figure out how to integrate this.

\int((y⁶ - 1)²/(2y³)) + 1)))^.5 dy

 

[Mark44]

Homework Statement

Determine the arc length of the function on the given interval

x = (y^4)/8 + 1/(4y^2) from 1 to 2

The arc length formula

\int (f'(x)² + 1)^.5 dx

The Attempt at a Solution

I used the arc length formula but don't know where to go from here. Usually these problems can't be done unless a form of cancellation takes place and I can't seem to find it. Below is what I input in the formula but I can not figure out how to integrate this.

\int((y⁶ - 1)²/(2y³)) + 1)))^.5 dy

It's helpful in this problem to use negative exponents instead of fractions.

(x')² + 1 = (1/4)y⁶ - 1/2 + (1/4)y^-6 + 1
= (1/4)y⁶ + 1/2 + (1/4)y^-6

This turns out to be a perfect square, so you can readily take its square root.

 

[MHrtz]

I factored and then took the square root and came up with this:

(y^3)/2 + 1/2y^3

 

[HallsofIvy]

Yes, that is correct. Now integrate.