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

1. Integrate term-by-term:

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

      $$\int 6 x^{2}\, dx = 6 \int x^{2}\, dx$$

      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: $2 x^{3}$

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

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

      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: $4 x^{2}$

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

2. Now simplify:

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

3. Add the constant of integration:

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

The answer is:

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