Integral of 16x^3 dx

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

   $$\int 16 x^{3}\, dx = 16 \int x^{3}\, dx$$

   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: $4 x^{4}$

2. Add the constant of integration:

   $$4 x^{4}+ \mathrm{constant}$$

The answer is:

$$4 x^{4}+ \mathrm{constant}$$