Integral of (8x^7+(6x)-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:

         $$\int 8 x^{7}\, dx = 8 \int x^{7}\, dx$$

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

            $$\int x^{7}\, dx = \frac{x^{8}}{8}$$

         So, the result is: $x^{8}$

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

         $$\int 6 x\, dx = 6 \int x\, dx$$

         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: $3 x^{2}$

      The result is: $x^{8} + 3 x^{2}$

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

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

   The result is: $x^{8} + 3 x^{2} - 3 x$

2. Now simplify:

   $$x \left(x^{7} + 3 x - 3\right)$$

3. Add the constant of integration:

   $$x \left(x^{7} + 3 x - 3\right)+ \mathrm{constant}$$

The answer is:

$$x \left(x^{7} + 3 x - 3\right)+ \mathrm{constant}$$