We need to evaluate the limit: $\lim_{x \to +\infty} \ln\left(\frac{(2x+1)^2}{2x^2+3x}\right)$.

1. Problem Description

We need to evaluate the limit:

$\lim_{x \to +\infty} \ln\left(\frac{(2x+1)^2}{2x^2+3x}\right)$.

2. Solution Steps

First, let's simplify the expression inside the logarithm:

$\frac{(2x+1)^2}{2x^2+3x} = \frac{4x^2 + 4x + 1}{2x^2+3x}$.

Now, we consider the limit as $x$ approaches infinity:

$\lim_{x \to +\infty} \frac{4x^2 + 4x + 1}{2x^2+3x}$.

To evaluate this limit, we can divide both the numerator and denominator by the highest power of $x$, which is $x^2$:

$\lim_{x \to +\infty} \frac{4 + \frac{4}{x} + \frac{1}{x^2}}{2 + \frac{3}{x}}$.

As $x \to +\infty$, the terms $\frac{4}{x}$, $\frac{1}{x^2}$, and $\frac{3}{x}$ approach

0. Therefore,

$\lim_{x \to +\infty} \frac{4 + \frac{4}{x} + \frac{1}{x^2}}{2 + \frac{3}{x}} = \frac{4 + 0 + 0}{2 + 0} = \frac{4}{2} = 2$.

Now, we can substitute this result back into the original limit:

$\lim_{x \to +\infty} \ln\left(\frac{(2x+1)^2}{2x^2+3x}\right) = \ln\left(\lim_{x \to +\infty} \frac{(2x+1)^2}{2x^2+3x}\right) = \ln(2)$.

3. Final Answer

$\ln(2)$