We will use one of the formulas for arc length:

`int_a^b sqrt(1+(dx/dy)^2) dy`

`x=(1/8)y^4+(1/4)y^(-2)`

`dx/dy = (1/2)y^3-(1/2)y^(-3)`

`(dx/dy)^2 = (1/4)y^6 - (1/2) + (1/4)y^(-6)`

`1+(dx/dy)^2 = (1/4)y^6 + 1/2 + (1/4)y^(-6) = ((1/2)y^3+(1/2)y^(-3))^2`

That is, we can rewrite our expression for `1+(dx/dy)^2` as a factored perfect square. This is convenient for...

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We will use one of the formulas for arc length:

`int_a^b sqrt(1+(dx/dy)^2) dy`

`x=(1/8)y^4+(1/4)y^(-2)`

`dx/dy = (1/2)y^3-(1/2)y^(-3)`

`(dx/dy)^2 = (1/4)y^6 - (1/2) + (1/4)y^(-6)`

`1+(dx/dy)^2 = (1/4)y^6 + 1/2 + (1/4)y^(-6) = ((1/2)y^3+(1/2)y^(-3))^2`

That is, we can rewrite our expression for `1+(dx/dy)^2` as a factored perfect square. This is convenient for taking the square root:

`sqrt(1+(dx/dy)^2) = (1/2)y^3+(1/2)y^(-3)`

Thus: `int_1^2 sqrt(1+(dx/dy)^2)dy = int_1^2 (1/2)y^3+(1/2)y^(-3) dy`

`=(1/8)y^4 -(1/4)y^(-2) |_1^2`

`=((2^4)/8-(2^(-2))/4)-((1^4)/8-(1^(-2))/4)`

`=(2-1/(16))-(1/8-1/4)=(33)/(16)`

**The length of the curve is** `33/16`