We are asked to evaluate the integral: $I = \int \frac{2x+1}{2x+3} dx$

1. Problem Description

We are asked to evaluate the integral:

$I = \int \frac{2x+1}{2x+3} dx$

2. Solution Steps

We can rewrite the fraction inside the integral as:

$\frac{2x+1}{2x+3} = \frac{(2x+3)-2}{2x+3} = \frac{2x+3}{2x+3} - \frac{2}{2x+3} = 1 - \frac{2}{2x+3}$

So, the integral becomes:

$I = \int (1 - \frac{2}{2x+3}) dx = \int 1 dx - \int \frac{2}{2x+3} dx = \int dx - \int \frac{2}{2x+3} dx$

The first integral is simply:

$\int dx = x + C_1$

For the second integral, let $u = 2x+3$, then $du = 2dx$. So, $dx = \frac{1}{2}du$. Thus,

$\int \frac{2}{2x+3} dx = \int \frac{2}{u} \frac{1}{2} du = \int \frac{1}{u} du = \ln|u| + C_2 = \ln|2x+3| + C_2$

Putting it together,

$I = x - \ln|2x+3| + C$, where $C = C_1 - C_2$.

3. Final Answer

$I = x - \ln|2x+3| + C$