Integral of x(x-3)² dx

1. Rewrite the integrand:

   $$x \left(x - 3\right)^{2} = x^{3} - 6 x^{2} + 9 x$$

2. Integrate term-by-term:

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

      $$\int x^{3}\, dx = \frac{x^{4}}{4}$$

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

      $$\int \left(- 6 x^{2}\right)\, dx = - 6 \int x^{2}\, dx$$

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

         $$\int x^{2}\, dx = \frac{x^{3}}{3}$$

      So, the result is: $- 2 x^{3}$

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

      $$\int 9 x\, dx = 9 \int x\, dx$$

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

         $$\int x\, dx = \frac{x^{2}}{2}$$

      So, the result is: $\frac{9 x^{2}}{2}$

   The result is: $\frac{x^{4}}{4} - 2 x^{3} + \frac{9 x^{2}}{2}$

3. Now simplify:

   $$\frac{x^{2} \left(x^{2} - 8 x + 18\right)}{4}$$

4. Add the constant of integration:

   $$\frac{x^{2} \left(x^{2} - 8 x + 18\right)}{4}+ \mathrm{constant}$$

The answer is:

$$\frac{x^{2} \left(x^{2} - 8 x + 18\right)}{4}+ \mathrm{constant}$$