Integral - Pls check my work

$\displaystyle \int_{-1}^{1} (\sqrt[3]{u} + 1)^{2} du$

This is what I did:

(FOIL)

$\displaystyle \int_{-1}^{1} ({u}^{2/3} + 2u^{1/3} + 1) du$

$\displaystyle \left[ \frac{3}{5}u^{5/3} + \frac{3}{2}u^{4/3} + u + C\right]_{-1}^{1}$

$\displaystyle \frac{3}{5}(1)^{5/3} + \frac{3}{2}(1)^{4/3} + 1 + C - \left(\frac{3}{5}(-1)^{5/3} + \frac{3}{2}(-1)^{4/3} - 1 + C \right)$

$\displaystyle \frac{3}{5} + 1 + \frac{3}{5} + 1$

$\displaystyle \frac{16}{5}$

But when I check it here: Function calculator it gives me what appears to be an imaginary number, it says the answer is: $\displaystyle 4.53546396068 + 1.80971485451I_{3}$. But I can't understand why, the negative ones should all work because they are never taken to an even root.