Integral of (3x^2+16x-2) dx

1. Integrate term-by-term:

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

      $$\int 3 x^{2}\, dx = 3 \int x^{2}\, dx$$

      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: $x^{3}$

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

      $$\int 16 x\, dx = 16 \int x\, dx$$

      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: $8 x^{2}$

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

      $$\int \left(\left(-1\right) 2\right)\, dx = - 2 x$$

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

2. Now simplify:

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

3. Add the constant of integration:

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

The answer is: