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

1. There are multiple ways to do this integral.

   Method #1

   1. Rewrite the integrand:

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

   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:

         $$\int \left(- \frac{1}{x + 2}\right)\, dx = - \int \frac{1}{x + 2}\, dx$$

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

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

   Method #2

   1. Rewrite the integrand:

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

   2. Integrate term-by-term:

      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:

            $$\int \left(- \frac{2}{x + 2}\right)\, dx = - 2 \int \frac{1}{x + 2}\, dx$$

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

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

2. Add the constant of integration:

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

The answer is:

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