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.
 A: Your derivative is wrong, and your squaring is wrong.


*

*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:
$$x^4 + \frac{1}{2} + \frac {1}{16}x^{-4} = (x^2)^2 + 2x^2\left(\frac{1}{4}x^{-2}\right) + \left(\frac{1}{4}x^{-2}\right)^2 = \left( x^2 + \frac{1}{4}x^{-2}\right)^2.$$
A: The derivative is $x^2-\frac{x^{-2}}{4}$. Its square is $x^4 -\frac{1}{2}+x^{-4}$. Add $1$, we get $x^4+\frac{1}{2}+x^{-4}$. The square root of this is $x^2+\frac{x^{-2}}{4}$. That should not be hard to integrate.
Remark: Examples of arclength are often artificial. This because for most functions $f(x)$, $\sqrt{1+(f'(x))^2}$ is something horrible that cannot be integrated in terms of elementary functions. For example, if instead of our function we had $\frac{x^3}{3}+\frac{x^{-2}}{5}$, we would end up with something that cannot be integrated in terms of elementary functions. For the same reason, any little mistake in differentiating or squaring usually leads to something one cannot integrate.  For that reason, I have mostly avoided putting such artificial examples on exams, since a minor slip can result in an excessive number of lost marks.
