hi can you guys help me with this please:

Evaluate the definite integral, if it exists. $\displaystyle \int_{-1}^{0} x^2(x^3+1)^8\! \, \mathrm{d}x.$

My teacher says there's a way that you don't have to bring back the original function, that you can just plug in for u. I got this far:

u+x^3+1

$\displaystyle \frac{du}{3}=x^2 dx $

$\displaystyle \frac{1}{3} \int_{-1}^{0} u^8\! \, \mathrm{d}u.$

where do I go from here to solve using only u?