I have a problem where I need to find the length of a given curve using integration. I've probably put about $2$ whole hours into this question, but I'm completely stumped as to solving it. Here's the question:
Find the arc length of $y = \dfrac32~x^{2/3}$ on the interval $1\le x\le8$.
I know to set up an integral on the bounds from $1$ to $8$, and write it as $1+\big(dy/dx\big)^2$ under a root, if that makes sense. I end up with the integral of $\sqrt{1 + x^{-2/3}}$ . I've tried everything I know of, but I simply cannot figure out how to process this integral. Can anybody offer an advice to solving this? Sorry if my formatting is bad, and thanks in advance.
This equals $x^{-1/3}\sqrt{x^{2/3}+1}$, so use the substitution $u=x^{2/3}$.