Integral of (9x-6)е^(6-2)x dx

1. Rewrite the integrand:

   $$x e^{4} \left(9 x - 6\right) = 9 x^{2} e^{4} - 6 x e^{4}$$

2. Integrate term-by-term:

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

      $$\int 9 x^{2} e^{4}\, dx = 9 e^{4} \int x^{2}\, dx$$

      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: $3 x^{3} e^{4}$

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

      $$\int \left(- 6 x e^{4}\right)\, dx = - 6 e^{4} \int x\, dx$$

      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: $- 3 x^{2} e^{4}$

   The result is: $3 x^{3} e^{4} - 3 x^{2} e^{4}$

3. Now simplify:

   $$3 x^{2} \left(x - 1\right) e^{4}$$

4. Add the constant of integration:

   $$3 x^{2} \left(x - 1\right) e^{4}+ \mathrm{constant}$$

The answer is:

$$3 x^{2} \left(x - 1\right) e^{4}+ \mathrm{constant}$$