Arc length of $y = \frac{x^3}{3} + \frac{1}{4x}$ Arc length of $y = \frac{x^3}{3} + \frac{1}{4x}$ over $1 \leq x \leq 2$
I know that the first thing I need to do is take the derivative.
$$y' = x^2 - 4x^{-2}$$
Then I take the integral on that range using the arc length formula.
$$\int_1^2 \sqrt{1 + (x^2-4x^{-2})^2}$$
$$(x^2-4x^{-2})^2 = -16x^{-4} - 8 + x^4$$
$$\int_1^2 \sqrt{-16x^{-4} - 7 + x^4 }$$
From here I have no idea how to factor this but I am pretty sure I must have messed up something before that.

Your derivative is wrong, and your squaring is wrong.


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*Your derivative is wrong:
$$\left(\frac{x^3}{3} +\frac{1}{4x}\right)' = \left(\frac{1}{3}x^3 + \frac{1}{4}x^{-1}\right)' = x^2 - \frac{1}{4}x^{-2} = x^2 - \frac{x^{-2}}{4}.$$

*You squared incorrectly: if you square $x^2-4x^{-2}$, you get:
$$(x^2-4x^{-2})^2 = x^4 - 8x^2x^{-2} + 16x^{-4} = x^4 - 8+16x^{-4}.$$
Note the plus sign on $16x^{-4}$; you have a minus sign.
If you square the correct function, you get
$$\left( x^2 - \frac{1}{4}x^{-2}\right)^2 = x^4 - \frac{1}{2} + \frac{1}{16}x^{-4}.$$
So the integral would be
$$\int_{1}^2 \sqrt{ x^4 + \frac{1}{2} + \frac{1}{16}x^{-4}}\,dx.$$
As for solving it, note that:
$$x^4 + \frac{1}{2} + \frac {1}{16}x^{-4} = (x^2)^2 + 2x^2\left(\frac{1}{4}x^{-2}\right) + \left(\frac{1}{4}x^{-2}\right)^2 = \left( x^2 + \frac{1}{4}x^{-2}\right)^2.$$
