Integral of (8x-4x^3)dx dx

1. Integrate term-by-term:

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

      $$\int \left(- 4 x^{3}\right)\, dx = - 4 \int x^{3}\, dx$$

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

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

      So, the result is: $- x^{4}$

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

      $$\int 8 x\, dx = 8 \int x\, dx$$

      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: $4 x^{2}$

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

2. Now simplify:

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

3. Add the constant of integration:

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

The answer is:

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