Integral of (6sqrt(x)+3x^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:

      $$\int 6 \sqrt{x}\, dx = 6 \int \sqrt{x}\, dx$$

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

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

      So, the result is: $4 x^{\frac{3}{2}}$

   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: $x^{3}$

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

2. Add the constant of integration:

   $$4 x^{\frac{3}{2}} + x^{3}+ \mathrm{constant}$$

The answer is:

$$4 x^{\frac{3}{2}} + x^{3}+ \mathrm{constant}$$