Integral of x^2+7x dx

1. Integrate term-by-term:

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

      $$\int x^{2}\, dx = \frac{x^{3}}{3}$$

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

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

      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: $\frac{7 x^{2}}{2}$

   The result is: $\frac{x^{3}}{3} + \frac{7 x^{2}}{2}$

2. Now simplify:

   $$\frac{x^{2} \cdot \left(2 x + 21\right)}{6}$$

3. Add the constant of integration:

   $$\frac{x^{2} \cdot \left(2 x + 21\right)}{6}+ \mathrm{constant}$$

The answer is:

$$\frac{x^{2} \cdot \left(2 x + 21\right)}{6}+ \mathrm{constant}$$