We need to evaluate the definite integral $I = \int_{0}^{2} (x + 1 + \frac{3}{x+1}) dx$.

1. Problem Description

We need to evaluate the definite integral $I = \int_{0}^{2} (x + 1 + \frac{3}{x+1}) dx$.

2. Solution Steps

First, we find the antiderivative of the integrand.

The antiderivative of $x$ is $\frac{x^2}{2}$.

The antiderivative of $1$ is $x$.

The antiderivative of $\frac{3}{x+1}$ is $3\ln|x+1|$.

Therefore, the antiderivative of $x + 1 + \frac{3}{x+1}$ is $\frac{x^2}{2} + x + 3\ln|x+1|$.

Next, we evaluate the definite integral by plugging in the limits of integration.

$I = \left[\frac{x^2}{2} + x + 3\ln|x+1|\right]_{0}^{2} = \left(\frac{2^2}{2} + 2 + 3\ln|2+1|\right) - \left(\frac{0^2}{2} + 0 + 3\ln|0+1|\right)$

$I = \left(\frac{4}{2} + 2 + 3\ln(3)\right) - \left(0 + 0 + 3\ln(1)\right)$

$I = (2 + 2 + 3\ln(3)) - (0)$

Since $\ln(1) = 0$.

$I = 4 + 3\ln(3)$

3. Final Answer

$4 + 3\ln(3)$