Integral of (4x³+6x) dx

1. Integrate term-by-term:

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

      $$\int 4 x^{3}\, dx = 4 \int x^{3}\, dx$$

      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: $x^{4}$

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

      $$\int 6 x\, dx = 6 \int x\, dx$$

      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: $3 x^{2}$

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

2. Now simplify:

   $$x^{2} \left(x^{2} + 3\right)$$

3. Add the constant of integration:

   $$x^{2} \left(x^{2} + 3\right)+ \mathrm{constant}$$

The answer is:

$$x^{2} \left(x^{2} + 3\right)+ \mathrm{constant}$$