The original question I assume was:
[tex]x=\frac{y^4}{8}+\frac{1}{4y^2},1\leq y\leq 2[/tex]
We'll use the formula for the arc length of a function which is the following:
[tex]L=\int ds\\
ds=\sqrt{1+\lef(\frac{dx}{dy} )^2}dy[/tex]
Now let's get the derivative of x:
[tex]\frac{dx}{dy}=\frac{4}{8}y^3-\frac{2}{4}\frac{1}{y^{3}}\\=\frac{1}{2}(y^3-\frac{1}{y^3})[/tex]
Alright now plug this into ds formula to get:
[tex]ds=\sqrt{1+\frac{1}{4}(y^3-\frac{1}{y^3})^2}[/tex]
Notice that:
[tex]1+\frac{1}{4}(y^3-\frac{1}{y^3})^2=1+\frac{1}{4}y^6-\frac{1}{2}+\frac{1}{4y^6}\\ =(\frac{1}{2}y^3+\frac{1}{2}y^{-3})^2
[/tex]
Now the integral will be more simple:
[tex]L=\int_1^2\sqrt{(\frac{1}{2}y^3+\frac{1}{2}y^{-3})^2}dy=\int_1^2\frac{1}{2}y^3+\frac{1}{2}y^{-3}dy\\=[\frac{1}{8}y^4-\frac{1}{4}y^{-2}]_2^1\\ =\frac{33}{16}[/tex]
