Integral of 32x^3-4x+ dx

1. Integrate term-by-term:

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

      $$\int 32 x^{3}\, dx = 32 \int x^{3}\, dx$$

      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: $8 x^{4}$

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

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

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

         $$\int 4 x\, dx = 4 \int x\, dx$$

         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: $2 x^{2}$

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

   The result is: $8 x^{4} - 2 x^{2}$

2. Now simplify:

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

3. Add the constant of integration:

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

The answer is: