Integral of (3x-1)(3x-1)dx dx

1. There are multiple ways to do this integral.

   Method #1

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

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

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

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

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

         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:

      $$\frac{\left(3 x - 1\right)^{3}}{9}$$

   Method #2

   1. Rewrite the integrand:

      $$\left(3 x - 1\right) \left(3 x - 1\right) = 9 x^{2} - 6 x + 1$$

   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:

         $$\int \left(- 6 x\right)\, dx = - 6 \int x\, dx$$

         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: $- 3 x^{2}$

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

         $$\int 1\, dx = x$$

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

2. Now simplify:

   $$\frac{\left(3 x - 1\right)^{3}}{9}$$

3. Add the constant of integration:

   $$\frac{\left(3 x - 1\right)^{3}}{9}+ \mathrm{constant}$$

The answer is:

$$\frac{\left(3 x - 1\right)^{3}}{9}+ \mathrm{constant}$$