Integral of x/(2x+1) dx

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

3. Add the constant of integration:

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

The answer is:

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