Integral of (8x^3-6x^2+3) dx

1. Integrate term-by-term:

   1. Integrate term-by-term:

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

         $$\int 8 x^{3}\, dx = 8 \int x^{3}\, dx$$

         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: $2 x^{4}$

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

         $$\int \left(- 6 x^{2}\right)\, dx = - 6 \int x^{2}\, dx$$

         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: $- 2 x^{3}$

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

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

      $$\int 3\, dx = 3 x$$

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

2. Now simplify:

   $$x \left(2 x^{3} - 2 x^{2} + 3\right)$$

3. Add the constant of integration:

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

The answer is:

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