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.