Integral of x^3-4x dx

1. 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:

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

      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: $- 2 x^{2}$

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

2. Now simplify:

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

3. Add the constant of integration:

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

The answer is:

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