Integral of (2x+1)/(x-2) dx

1. There are multiple ways to do this integral.

   Method #1

   1. Let $u = 2 x$.

      Then let $du = 2 dx$ and substitute $du$:

      $$\int \frac{u + 1}{u - 4}\, du$$

      1. Let $u = u - 4$.

         Then let $du = du$ and substitute $du$:

         $$\int \frac{u + 5}{u}\, du$$

         1. Rewrite the integrand:

            $$\frac{u + 5}{u} = 1 + \frac{5}{u}$$

         2. Integrate term-by-term:

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

               $$\int 1\, du = u$$

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

               $$\int \frac{5}{u}\, du = 5 \int \frac{1}{u}\, du$$

               1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$.

               So, the result is: $5 \log{\left(u \right)}$

            The result is: $u + 5 \log{\left(u \right)}$

         Now substitute $u$ back in:

         $$u + 5 \log{\left(u - 4 \right)} - 4$$

      Now substitute $u$ back in:

      $$2 x + 5 \log{\left(2 x - 4 \right)} - 4$$

   Method #2

   1. Rewrite the integrand:

      $$\frac{2 x + 1}{x - 2} = 2 + \frac{5}{x - 2}$$

   2. Integrate term-by-term:

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

         $$\int 2\, dx = 2 x$$

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

         $$\int \frac{5}{x - 2}\, dx = 5 \int \frac{1}{x - 2}\, dx$$

         1. Let $u = x - 2$.

            Then let $du = dx$ and substitute $du$:

            $$\int \frac{1}{u}\, du$$

            1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$.

            Now substitute $u$ back in:

            $$\log{\left(x - 2 \right)}$$

         So, the result is: $5 \log{\left(x - 2 \right)}$

      The result is: $2 x + 5 \log{\left(x - 2 \right)}$

   Method #3

   1. Rewrite the integrand:

      $$\frac{2 x + 1}{x - 2} = \frac{2 x}{x - 2} + \frac{1}{x - 2}$$

   2. Integrate term-by-term:

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

         $$\int \frac{2 x}{x - 2}\, dx = 2 \int \frac{x}{x - 2}\, dx$$

         1. Rewrite the integrand:

            $$\frac{x}{x - 2} = 1 + \frac{2}{x - 2}$$

         2. Integrate term-by-term:

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

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

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

               $$\int \frac{2}{x - 2}\, dx = 2 \int \frac{1}{x - 2}\, dx$$

               1. Let $u = x - 2$.

                  Then let $du = dx$ and substitute $du$:

                  $$\int \frac{1}{u}\, du$$

                  1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$.

                  Now substitute $u$ back in:

                  $$\log{\left(x - 2 \right)}$$

               So, the result is: $2 \log{\left(x - 2 \right)}$

            The result is: $x + 2 \log{\left(x - 2 \right)}$

         So, the result is: $2 x + 4 \log{\left(x - 2 \right)}$

      1. Let $u = x - 2$.

         Then let $du = dx$ and substitute $du$:

         $$\int \frac{1}{u}\, du$$

         1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$.

         Now substitute $u$ back in:

         $$\log{\left(x - 2 \right)}$$

      The result is: $2 x + 4 \log{\left(x - 2 \right)} + \log{\left(x - 2 \right)}$

2. Add the constant of integration:

   $$2 x + 5 \log{\left(2 x - 4 \right)} - 4+ \mathrm{constant}$$

The answer is:

$$2 x + 5 \log{\left(2 x - 4 \right)} - 4+ \mathrm{constant}$$