Integral of x^3-4x+7 dx

1. Integrate term-by-term:

   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:

         $$\int \left(- 4 x\right)\, dx = - 4 \int x\, dx$$

         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: $- 2 x^{2}$

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

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

      $$\int 7\, dx = 7 x$$

   The result is: $\frac{x^{4}}{4} - 2 x^{2} + 7 x$

2. Now simplify:

   $$\frac{x \left(x^{3} - 8 x + 28\right)}{4}$$

3. Add the constant of integration:

   $$\frac{x \left(x^{3} - 8 x + 28\right)}{4}+ \mathrm{constant}$$

The answer is:

$$\frac{x \left(x^{3} - 8 x + 28\right)}{4}+ \mathrm{constant}$$