# Find the exact length of the curve x=\frac{y^4}{8}+\frac{1}{4y^2}\quad1\leq y\leq2

Find the exact length of the curve
$x=\frac{{y}^{4}}{8}+\frac{1}{4{y}^{2}}\phantom{\rule{1em}{0ex}}1\le y\le 2$
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