Integral of x^4(x-1) dx

1. Rewrite the integrand:

   $$x^{4} \left(x - 1\right) = x^{5} - x^{4}$$

2. Integrate term-by-term:

   1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$:

      $$\int x^{5}\, dx = \frac{x^{6}}{6}$$

   1. The integral of a constant times a function is the constant times the integral of the function:

      $$\int \left(- x^{4}\right)\, dx = - \int x^{4}\, dx$$

      1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$:

         $$\int x^{4}\, dx = \frac{x^{5}}{5}$$

      So, the result is: $- \frac{x^{5}}{5}$

   The result is: $\frac{x^{6}}{6} - \frac{x^{5}}{5}$

3. Now simplify:

   $$\frac{x^{5} \cdot \left(5 x - 6\right)}{30}$$

4. Add the constant of integration:

   $$\frac{x^{5} \cdot \left(5 x - 6\right)}{30}+ \mathrm{constant}$$

The answer is:

$$\frac{x^{5} \cdot \left(5 x - 6\right)}{30}+ \mathrm{constant}$$