Integral of 8-x^3 dx

1. Integrate term-by-term:

   1. The integral of a constant is the constant times the variable of integration:

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

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

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

      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: $- \frac{x^{4}}{4}$

   The result is: $- \frac{x^{4}}{4} + 8 x$

2. Now simplify:

   $$\frac{x \left(32 - x^{3}\right)}{4}$$

3. Add the constant of integration:

   $$\frac{x \left(32 - x^{3}\right)}{4}+ \mathrm{constant}$$

The answer is:

$$\frac{x \left(32 - x^{3}\right)}{4}+ \mathrm{constant}$$