Integral of (x-8)^2 dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = x - 8$.

      Then let $du = dx$ and substitute $du$:

      $$\int u^{2}\, du$$

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

         $$\int u^{2}\, du = \frac{u^{3}}{3}$$

      Now substitute $u$ back in:

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

   Method #2

   1. Rewrite the integrand:

      $$\left(x - 8\right)^{2} = x^{2} - 16 x + 64$$

   2. 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:

         $$\int \left(- 16 x\right)\, dx = - 16 \int x\, dx$$

         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: $- 8 x^{2}$

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

         $$\int 64\, dx = 64 x$$

      The result is: $\frac{x^{3}}{3} - 8 x^{2} + 64 x$

2. Now simplify:

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

3. Add the constant of integration:

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

The answer is:

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