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

1. There are multiple ways to do this integral.

   Method #1

   1. Rewrite the integrand:

      $$\frac{x + 1}{2 x + 1} = \frac{1}{2} + \frac{1}{2 \left(2 x + 1\right)}$$

   2. Integrate term-by-term:

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

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

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

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

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

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

            $$\int \frac{1}{2 u}\, du$$

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

               $$\int \frac{1}{u}\, du = \frac{\int \frac{1}{u}\, du}{2}$$

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

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

            Now substitute $u$ back in:

            $$\frac{\log{\left(2 x + 1 \right)}}{2}$$

         So, the result is: $\frac{\log{\left(2 x + 1 \right)}}{4}$

      The result is: $\frac{x}{2} + \frac{\log{\left(2 x + 1 \right)}}{4}$

   Method #2

   1. Rewrite the integrand:

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

   2. Integrate term-by-term:

      1. Rewrite the integrand:

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

      2. Integrate term-by-term:

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

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

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

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

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

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

               $$\int \frac{1}{2 u}\, du$$

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

                  $$\int \frac{1}{u}\, du = \frac{\int \frac{1}{u}\, du}{2}$$

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

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

               Now substitute $u$ back in:

               $$\frac{\log{\left(2 x + 1 \right)}}{2}$$

            So, the result is: $- \frac{\log{\left(2 x + 1 \right)}}{4}$

         The result is: $\frac{x}{2} - \frac{\log{\left(2 x + 1 \right)}}{4}$

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

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

         $$\int \frac{1}{2 u}\, du$$

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

            $$\int \frac{1}{u}\, du = \frac{\int \frac{1}{u}\, du}{2}$$

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

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

         Now substitute $u$ back in:

         $$\frac{\log{\left(2 x + 1 \right)}}{2}$$

      The result is: $\frac{x}{2} + \frac{\log{\left(2 x + 1 \right)}}{2} - \frac{\log{\left(2 x + 1 \right)}}{4}$

2. Add the constant of integration:

   $$\frac{x}{2} + \frac{\log{\left(2 x + 1 \right)}}{4}+ \mathrm{constant}$$

The answer is:

$$\frac{x}{2} + \frac{\log{\left(2 x + 1 \right)}}{4}+ \mathrm{constant}$$