Integral of (3t-1)^3dt dx

1. There are multiple ways to do this integral.

   Method #1

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

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

      $$\int \frac{u^{3}}{9}\, du$$

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

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

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

            $$\int u^{3}\, du = \frac{u^{4}}{4}$$

         So, the result is: $\frac{u^{4}}{12}$

      Now substitute $u$ back in:

      $$\frac{\left(3 t - 1\right)^{4}}{12}$$

   Method #2

   1. Rewrite the integrand:

      $$\left(3 t - 1\right)^{3} \cdot 1 = 27 t^{3} - 27 t^{2} + 9 t - 1$$

   2. Integrate term-by-term:

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

         $$\int 27 t^{3}\, dt = 27 \int t^{3}\, dt$$

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

            $$\int t^{3}\, dt = \frac{t^{4}}{4}$$

         So, the result is: $\frac{27 t^{4}}{4}$

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

         $$\int \left(- 27 t^{2}\right)\, dt = - 27 \int t^{2}\, dt$$

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

            $$\int t^{2}\, dt = \frac{t^{3}}{3}$$

         So, the result is: $- 9 t^{3}$

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

         $$\int 9 t\, dt = 9 \int t\, dt$$

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

            $$\int t\, dt = \frac{t^{2}}{2}$$

         So, the result is: $\frac{9 t^{2}}{2}$

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

         $$\int \left(-1\right)\, dt = - t$$

      The result is: $\frac{27 t^{4}}{4} - 9 t^{3} + \frac{9 t^{2}}{2} - t$

2. Now simplify:

   $$\frac{\left(3 t - 1\right)^{4}}{12}$$

3. Add the constant of integration:

   $$\frac{\left(3 t - 1\right)^{4}}{12}+ \mathrm{constant}$$

The answer is:

$$\frac{\left(3 t - 1\right)^{4}}{12}+ \mathrm{constant}$$