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$y = \frac{x^4}{16} + \frac{1}{2x^2}$, $1 \leq x \leq 2$

$\frac{dy}{dx} = \frac{x^3}{4} - \frac{1}{x^3}dx$ $$\begin{align} L &= \int_{1}^{2} \sqrt{1 + \bigg[ \frac{x^3}{4} - \frac{1}{x^3} \bigg]^2}dx \\ &= \int_{1}^{2} \sqrt{1 + \bigg[ \frac{x^6-4}{4x^3}\bigg]^2}dx \\ &= \int_{1}^{2} \sqrt{\frac{(4x^3)^2 + (x^6-4)^2}{(4x^3)^2}}dx \end{align}$$

I tried splitting the $\sqrt{~~}$ into the numerator and denominator, but that didn't really get me anywhere and I am stuck. Can someone give me a hint in the right direction possibly?

edit: continuing on...

$$\begin{align} L &= \int_{1}^{2} \sqrt{\frac{(x^6+4)^2}{(4x^3)^2}}dx \\ &= \int_{1}^{2} \frac{(x^6+4)}{(4x^3)} \\ &= 4 \int_{1}^{2} \frac{x^6+4}{x^3}dx \\ &= 4\int_{1}^{2} \bigg(x^3 + \frac{4}{x^3}\bigg)dx \\ &= 4 \bigg[ \frac{1}{4}x^4 - \frac{2}{x^2} \bigg]_{1}^{2} \\ &= 4 \bigg[ (4-\frac{1}{2}) - (\frac{1}{4} - 2)\bigg] \\ &= 21 \end{align}$$

However wolphram alpha says the answer is actually $\frac{21}{16}$.

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$$(4x^3)^2 + (x^6-4)^2$$ $$=16x^6 + x^{12} + 16 - 8x^6$$ $$=(x^6+4)^2$$

So that square root will open up straight away.

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    $\begingroup$ ah, I factored $(x^6+4)^2$ wrong, did it too hastily. I see my error, thank you! $\endgroup$ – Evan Kim May 8 '19 at 21:22
  • $\begingroup$ Yeah sure. No problem! $\endgroup$ – Vizag May 8 '19 at 21:23
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    $\begingroup$ Yeah why did you send that $4$ in the numerator? $\endgroup$ – Vizag May 8 '19 at 22:00
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    $\begingroup$ $= \int_{1}^{2} \frac{(x^6+4)}{(4x^3)}$... your error is after this line when you mistakenly take $4$ in the numerator $\endgroup$ – Vizag May 8 '19 at 22:02
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    $\begingroup$ Yeah sure. Please mark the answer as accepted. :) $\endgroup$ – Vizag May 12 '19 at 22:58

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