Integral of (2x-1)^4 dx

1. There are multiple ways to do this integral.

   Method #1

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

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

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

      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:

      $$\frac{\left(2 x - 1\right)^{5}}{10}$$

   Method #2

   1. Rewrite the integrand:

      $$\left(2 x - 1\right)^{4} = 16 x^{4} - 32 x^{3} + 24 x^{2} - 8 x + 1$$

   2. Integrate term-by-term:

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

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

         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:

         $$\int \left(- 32 x^{3}\right)\, dx = - 32 \int x^{3}\, dx$$

         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: $- 8 x^{4}$

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

         $$\int 24 x^{2}\, dx = 24 \int x^{2}\, dx$$

         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: $8 x^{3}$

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

         $$\int \left(- 8 x\right)\, dx = - 8 \int x\, dx$$

         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: $- 4 x^{2}$

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

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

      The result is: $\frac{16 x^{5}}{5} - 8 x^{4} + 8 x^{3} - 4 x^{2} + x$

2. Now simplify:

   $$\frac{\left(2 x - 1\right)^{5}}{10}$$

3. Add the constant of integration:

   $$\frac{\left(2 x - 1\right)^{5}}{10}+ \mathrm{constant}$$

The answer is: