Integral of (10x+3x^3-2) dx

1. Integrate term-by-term:

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

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

      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: $\frac{3 x^{4}}{4}$

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

      $$\int 10 x\, dx = 10 \int x\, dx$$

      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: $5 x^{2}$

   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: $\frac{3 x^{4}}{4} + 5 x^{2} - 2 x$

2. Now simplify:

   $$\frac{x \left(3 x^{3} + 20 x - 8\right)}{4}$$

3. Add the constant of integration:

   $$\frac{x \left(3 x^{3} + 20 x - 8\right)}{4}+ \mathrm{constant}$$

The answer is:

$$\frac{x \left(3 x^{3} + 20 x - 8\right)}{4}+ \mathrm{constant}$$