Integral of x^3/4dx dx

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

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

   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: $\frac{x^{4}}{16}$

2. Add the constant of integration:

   $$\frac{x^{4}}{16}+ \mathrm{constant}$$

The answer is:

$$\frac{x^{4}}{16}+ \mathrm{constant}$$