Integral of f(x)=(4x-5)^6 dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = 4 x - 5$.

      Then let $du = 4 dx$ and substitute $\frac{du}{4}$:

      $$\int \frac{u^{6}}{4}\, du$$

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

         $$\int u^{6}\, du = \frac{\int u^{6}\, du}{4}$$

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

            $$\int u^{6}\, du = \frac{u^{7}}{7}$$

         So, the result is: $\frac{u^{7}}{28}$

      Now substitute $u$ back in:

      $$\frac{\left(4 x - 5\right)^{7}}{28}$$

   Method #2

   1. Rewrite the integrand:

      $$\left(4 x - 5\right)^{6} = 4096 x^{6} - 30720 x^{5} + 96000 x^{4} - 160000 x^{3} + 150000 x^{2} - 75000 x + 15625$$

   2. Integrate term-by-term:

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

         $$\int 4096 x^{6}\, dx = 4096 \int x^{6}\, dx$$

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

            $$\int x^{6}\, dx = \frac{x^{7}}{7}$$

         So, the result is: $\frac{4096 x^{7}}{7}$

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

         $$\int \left(- 30720 x^{5}\right)\, dx = - 30720 \int x^{5}\, dx$$

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

            $$\int x^{5}\, dx = \frac{x^{6}}{6}$$

         So, the result is: $- 5120 x^{6}$

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

         $$\int 96000 x^{4}\, dx = 96000 \int x^{4}\, dx$$

         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: $19200 x^{5}$

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

         $$\int \left(- 160000 x^{3}\right)\, dx = - 160000 \int x^{3}\, dx$$

         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: $- 40000 x^{4}$

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

         $$\int 150000 x^{2}\, dx = 150000 \int x^{2}\, dx$$

         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: $50000 x^{3}$

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

         $$\int \left(- 75000 x\right)\, dx = - 75000 \int x\, dx$$

         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: $- 37500 x^{2}$

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

         $$\int 15625\, dx = 15625 x$$

      The result is: $\frac{4096 x^{7}}{7} - 5120 x^{6} + 19200 x^{5} - 40000 x^{4} + 50000 x^{3} - 37500 x^{2} + 15625 x$

2. Now simplify:

   $$\frac{\left(4 x - 5\right)^{7}}{28}$$

3. Add the constant of integration:

   $$\frac{\left(4 x - 5\right)^{7}}{28}+ \mathrm{constant}$$

The answer is:

$$\frac{\left(4 x - 5\right)^{7}}{28}+ \mathrm{constant}$$