Chapter 13.2, Problem 106E

To determine

The reason behind the following calculation of the integral being wrong and what will be the correct solution,

$\begin{array}{c}{{\displaystyle \int x^2\left(x^3-1\right)}}^{2}dx={\displaystyle \int x^2{u}^2\frac{1}{3x^2}du}\\ =\frac{1}{3}{u}^2+C\\ =\frac{1}{3}{\left(x^3-1\right)}^2+C\end{array}$

Where, the substitution is $u=x^3-1,\frac{du}{dx}=3x^2\text{and}dx=\frac{1}{3x^2}du$.