Each time try solve this, the answer i get doesn't match the answer in the back of the book. These are the steps i take to solve this integration: (Please correct me where i'm going wrong!)

$\displaystyle \int x^2(x^3+1)^4 dx$

let $\displaystyle u = x^3 + 1$

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

$\displaystyle du = 3x^2 dx$

$\displaystyle dx = \frac{1}{3x^2} du$

$\displaystyle \int x^2u^4 dx$

$\displaystyle \int x^2u^4 \frac{1}{3x^2} du$

$\displaystyle \frac{x^3}{3} . \frac{u^5}{5} + c$

$\displaystyle \frac{x^3(u^5)}{15} + c$

$\displaystyle \frac{x^3(x^3 + 1)^5}{15} + c$

The answer in the back of the book is: $\displaystyle \frac{1}{15}(x^3 + 1)^5 + c$

This isn't urgent, however I would love to see where I'm going wrong here. Thanks