Integral of (3-2x)⁴dx dx

1. There are multiple ways to do this integral.

   Method #1

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

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

      $$\int \left(- \frac{u^{4}}{2}\right)\, du$$

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

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

         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: $- \frac{u^{5}}{10}$

      Now substitute $u$ back in:

      $$- \frac{\left(3 - 2 x\right)^{5}}{10}$$

   Method #2

   1. Rewrite the integrand:

      $$\left(3 - 2 x\right)^{4} = 16 x^{4} - 96 x^{3} + 216 x^{2} - 216 x + 81$$

   2. Integrate term-by-term:

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

         $$\int 16 x^{4}\, dx = 16 \int x^{4}\, dx$$

         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: $\frac{16 x^{5}}{5}$

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

         $$\int \left(- 96 x^{3}\right)\, dx = - 96 \int x^{3}\, dx$$

         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: $- 24 x^{4}$

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

         $$\int 216 x^{2}\, dx = 216 \int x^{2}\, dx$$

         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: $72 x^{3}$

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

         $$\int \left(- 216 x\right)\, dx = - 216 \int x\, dx$$

         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: $- 108 x^{2}$

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

         $$\int 81\, dx = 81 x$$

      The result is: $\frac{16 x^{5}}{5} - 24 x^{4} + 72 x^{3} - 108 x^{2} + 81 x$

2. Now simplify:

   $$\frac{\left(2 x - 3\right)^{5}}{10}$$

3. Add the constant of integration:

   $$\frac{\left(2 x - 3\right)^{5}}{10}+ \mathrm{constant}$$

The answer is: