# Find the exact length of the curve. x=\frac{y^4}{8}+\frac{1}{4y^2}, 1\leq y\leq2

Find the exact length of the curve
$$\displaystyle{x}={\frac{{{y}^{{4}}}}{{{8}}}}+{\frac{{{1}}}{{{4}{y}^{{2}}}}}\quad{1}\leq{y}\leq{2}$$

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toroztatG
You need to apply the next formula