Integral of (15-3x)^2 dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = 15 - 3 x$.

      Then let $du = - 3 dx$ and substitute $- \frac{du}{3}$:

      $$\int \frac{u^{2}}{9}\, du$$

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

         $$\int \left(- \frac{u^{2}}{3}\right)\, du = - \frac{\int u^{2}\, du}{3}$$

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

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

         So, the result is: $- \frac{u^{3}}{9}$

      Now substitute $u$ back in:

      $$- \frac{\left(15 - 3 x\right)^{3}}{9}$$

   Method #2

   1. Rewrite the integrand:

      $$\left(15 - 3 x\right)^{2} = 9 x^{2} - 90 x + 225$$

   2. Integrate term-by-term:

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

         $$\int 9 x^{2}\, dx = 9 \int x^{2}\, dx$$

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

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

         So, the result is: $3 x^{3}$

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

         $$\int \left(- 90 x\right)\, dx = - 90 \int x\, dx$$

         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: $- 45 x^{2}$

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

         $$\int 225\, dx = 225 x$$

      The result is: $3 x^{3} - 45 x^{2} + 225 x$

2. Now simplify:

   $$3 \left(x - 5\right)^{3}$$

3. Add the constant of integration:

   $$3 \left(x - 5\right)^{3}+ \mathrm{constant}$$

The answer is: