Integral of (3x+1)^4 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 $\frac{du}{3}$:

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

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

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

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

            $$\int u^{4}\, du = \frac{u^{5}}{5}$$

         So, the result is: $\frac{u^{5}}{15}$

      Now substitute $u$ back in:

      $$\frac{\left(3 x + 1\right)^{5}}{15}$$

   Method #2

   1. Rewrite the integrand:

      $$\left(3 x + 1\right)^{4} = 81 x^{4} + 108 x^{3} + 54 x^{2} + 12 x + 1$$

   2. Integrate term-by-term:

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

         $$\int 81 x^{4}\, dx = 81 \int x^{4}\, dx$$

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

            $$\int x^{4}\, dx = \frac{x^{5}}{5}$$

         So, the result is: $\frac{81 x^{5}}{5}$

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

         $$\int 108 x^{3}\, dx = 108 \int x^{3}\, dx$$

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

            $$\int x^{3}\, dx = \frac{x^{4}}{4}$$

         So, the result is: $27 x^{4}$

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

         $$\int 54 x^{2}\, dx = 54 \int x^{2}\, dx$$

         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: $18 x^{3}$

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

         $$\int 12 x\, dx = 12 \int x\, dx$$

         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:

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

      The result is: $\frac{81 x^{5}}{5} + 27 x^{4} + 18 x^{3} + 6 x^{2} + x$

2. Now simplify:

   $$\frac{\left(3 x + 1\right)^{5}}{15}$$

3. Add the constant of integration:

   $$\frac{\left(3 x + 1\right)^{5}}{15}+ \mathrm{constant}$$

The answer is:

$$\frac{\left(3 x + 1\right)^{5}}{15}+ \mathrm{constant}$$