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If [tex] y = \frac{x^{3}}{6} + \frac{1}{2x}\ [/tex] and [tex] \frac{1}{2}\leq x\leq 1 [/tex]. Find the arc length.

So [tex] \frac{dy}{dx} = \frac{x^{2}}{2} - \frac{1}{2x^{2}} [/tex]. So I got [tex] \frac{1}{2} \int^{1}_{\frac{1}{2}} \sqrt{2+x^{4} + x^{-4}} dx [/tex]. How would you evaulate this?

Thanks

So [tex] \frac{dy}{dx} = \frac{x^{2}}{2} - \frac{1}{2x^{2}} [/tex]. So I got [tex] \frac{1}{2} \int^{1}_{\frac{1}{2}} \sqrt{2+x^{4} + x^{-4}} dx [/tex]. How would you evaulate this?

Thanks

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