Arc length of $y = x^{2/3}$ 
Find the arc length of $y=x^{2/3}$ from $x=-1$ to $x=8$. 

I have tried applying the arc length formula but for some reason I keep getting $7.63$, but the answer is $10.51$.
 A: The complex outcome of Maazul seems to be caused by the definition of $x^{1/3}$ for $x < 0$. Many computer algebra packages do not take it to be the "natural choice" $-|x|^{1/3}$ (but one of the other complex roots), while it is almost certain that this is what is intended.
Therefore, the following calculation (which should be reminiscent of what OP did) is tempting:
$$\begin{align}
\int_{-1}^8 \sqrt{1+\frac49x^{-2/3}}\,\mathrm dx &=\int_{-1}^8\sqrt{\frac{x^{2/3}+\frac49}{x^{2/3}}}\,\mathrm dx\\{\text{tentative}}&= \int_{-1}^8 \frac{\sqrt{x^{2/3}+\frac49}}{x^{1/3}}\,\mathrm dx \\&= \left.\left(\frac49+x^{2/3}\right)^{3/2}\right|_{x=-1}^8 = \left(\frac{40}9\right)^{3/2}-\left(\frac{13}9\right)^{3/2} \approx 7.63
\end{align}$$
but $\sqrt{x^{2/3}} \ne x^{1/3}$ for $x < 0$, at which point a correcting minus sign is required. 
Therefore, we have to split the integral into two parts: $\int_{-1}^0$ and $\int_0^8$ (which will have different primitives). Since this is a homework assignment, I'll stop here.

Addendum after seeing OP evaluates using substitution: Similar problems to above arise when using $u = x^{2/3}$ because this $u$ is always positive. Therefore, splitting the interval of integration is again necessary. Then both integrals can be solved using (suitable variants of) this substitution.
A: $\displaystyle y=x^{2/3}$
$\displaystyle y'=\frac{2}{3}x^{-1/3}$
$\displaystyle L=\int_{-1}^{8} ds$
$\displaystyle L=\int_{-1}^{8} \sqrt{dx^2+dy^2}$
$\displaystyle L=\int_{-1}^{8} \sqrt{1+y'^2}dx$
$\displaystyle L=\int_{-1}^{8} \sqrt{1+\frac{4}{9}x^{-2/3}}dx\approx10+0.51i$
