Integral of (2x-3)^2 dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = 2 x - 3$.

      Then let $du = 2 dx$ and substitute $\frac{du}{2}$:

      $$\int \frac{u^{2}}{4}\, du$$

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

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

         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: $\frac{u^{3}}{6}$

      Now substitute $u$ back in:

      $$\frac{\left(2 x - 3\right)^{3}}{6}$$

   Method #2

   1. Rewrite the integrand:

      $$\left(2 x - 3\right)^{2} = 4 x^{2} - 12 x + 9$$

   2. Integrate term-by-term:

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

         $$\int 4 x^{2}\, dx = 4 \int x^{2}\, dx$$

         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: $\frac{4 x^{3}}{3}$

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

         $$\int \left(- 12 x\right)\, dx = - 12 \int x\, dx$$

         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: $- 6 x^{2}$

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

         $$\int 9\, dx = 9 x$$

      The result is: $\frac{4 x^{3}}{3} - 6 x^{2} + 9 x$

2. Now simplify:

   $$\frac{\left(2 x - 3\right)^{3}}{6}$$

3. Add the constant of integration:

   $$\frac{\left(2 x - 3\right)^{3}}{6}+ \mathrm{constant}$$

The answer is:

$$\frac{\left(2 x - 3\right)^{3}}{6}+ \mathrm{constant}$$