Integral of x^2-8x+16 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:

      $$\int \left(- 8 x\right)\, dx = - \int 8 x\, dx$$

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

         $$\int 8 x\, dx = 8 \int x\, dx$$

         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: $4 x^{2}$

      So, the result is: $- 4 x^{2}$

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

      $$\int 16\, dx = 16 x$$

   The result is: $\frac{x^{3}}{3} - 4 x^{2} + 16 x$

2. Now simplify:

   $$\frac{x \left(x^{2} - 12 x + 48\right)}{3}$$

3. Add the constant of integration:

   $$\frac{x \left(x^{2} - 12 x + 48\right)}{3}+ \mathrm{constant}$$

The answer is:

$$\frac{x \left(x^{2} - 12 x + 48\right)}{3}+ \mathrm{constant}$$