Integral of 5+3x-2x^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:

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

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

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

         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: $\frac{2 x^{3}}{3}$

      So, the result is: $- \frac{2 x^{3}}{3}$

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

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

      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: $\frac{3 x^{2}}{2}$

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

      $$\int 5\, dx = 5 x$$

   The result is: $- \frac{2 x^{3}}{3} + \frac{3 x^{2}}{2} + 5 x$

2. Now simplify:

   $$\frac{x \left(- 4 x^{2} + 9 x + 30\right)}{6}$$

3. Add the constant of integration:

   $$\frac{x \left(- 4 x^{2} + 9 x + 30\right)}{6}+ \mathrm{constant}$$

The answer is:

$$\frac{x \left(- 4 x^{2} + 9 x + 30\right)}{6}+ \mathrm{constant}$$