We are given the equation $\frac{3x+1}{x+2} = 1 + \frac{x+1}{x-2}$ and we want to solve for $x$.

1. Problem Description

We are given the equation $\frac{3x+1}{x+2} = 1 + \frac{x+1}{x-2}$ and we want to solve for $x$.

2. Solution Steps

First, we want to isolate $x$ by performing algebraic manipulations to the equation.

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

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

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

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

Now, we multiply both sides by $(x+2)(x-2)$:

$(3x+1)(x-2) = (2x-1)(x+2)$

Expanding both sides, we get:

$3x^2 - 6x + x - 2 = 2x^2 + 4x - x - 2$

$3x^2 - 5x - 2 = 2x^2 + 3x - 2$

Subtract $2x^2 + 3x - 2$ from both sides:

$3x^2 - 5x - 2 - (2x^2 + 3x - 2) = 0$

$3x^2 - 5x - 2 - 2x^2 - 3x + 2 = 0$

$x^2 - 8x = 0$

$x(x-8) = 0$

This implies that $x=0$ or $x-8=0$, which means $x=8$.

Now, we check the solutions to make sure that the denominators are not zero.

If $x=0$, then the denominators are $x+2 = 0+2 = 2$ and $x-2 = 0-2 = -2$, which are both non-zero. So $x=0$ is a valid solution.

If $x=8$, then the denominators are $x+2 = 8+2 = 10$ and $x-2 = 8-2 = 6$, which are both non-zero. So $x=8$ is a valid solution.

3. Final Answer

The solutions are $x=0$ and $x=8$.

Final Answer: The final answer is $\boxed{0, 8}$