Integral of x^3+3x-1 dx

1. Integrate term-by-term:

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

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

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

      $$\int 3 x\, dx = 3 \int x\, dx$$

      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: $\frac{3 x^{2}}{2}$

   1. The integral of a constant is the constant times the variable of integration:

      $$\int \left(\left(-1\right) 1\right)\, dx = - x$$

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

2. Now simplify:

   $$\frac{x \left(x^{3} + 6 x - 4\right)}{4}$$

3. Add the constant of integration:

   $$\frac{x \left(x^{3} + 6 x - 4\right)}{4}+ \mathrm{constant}$$

The answer is:

$$\frac{x \left(x^{3} + 6 x - 4\right)}{4}+ \mathrm{constant}$$