To solve the equation

\[ \frac{x + 1}{x} - \frac{3x}{x(x+4)} = \frac{x - 2}{x + 4}, \]

we will first simplify each term and then solve for \( x \).

1. Simplify the left-hand side:

\[ \frac{x + 1}{x} - \frac{3x}{x(x+4)} \]

The first term simplifies to:

\[ \frac{x + 1}{x} = \frac{x}{x} + \frac{1}{x} = 1 + \frac{1}{x} \]

The second term simplifies to:

\[ \frac{3x}{x(x+4)} = \frac{3}{x+4} \]

So the left-hand side becomes:

\[ 1 + \frac{1}{x} - \frac{3}{x+4} \]

2. Set the simplified left-hand side equal to the right-hand side:

\[ 1 + \frac{1}{x} - \frac{3}{x+4} = \frac{x - 2}{x + 4} \]

3. Combine the fractions on the left-hand side:

To combine the fractions, find a common denominator, which is \( x(x+4) \):

\[ 1 + \frac{1}{x} - \frac{3}{x+4} = 1 + \frac{(x+4) - 3x}{x(x+4)} = 1 + \frac{x + 4 - 3x}{x(x+4)} = 1 + \frac{-2x + 4}{x(x+4)} \]

So the equation becomes:

\[ 1 + \frac{-2x + 4}{x(x+4)} = \frac{x - 2}{x + 4} \]

4. Combine the terms on the left-hand side:

\[ 1 + \frac{-2x + 4}{x(x+4)} = \frac{x(x+4) + (-2x + 4)}{x(x+4)} = \frac{x^2 + 4x - 2x + 4}{x(x+4)} = \frac{x^2 + 2x + 4}{x(x+4)} \]

So the equation becomes:

\[ \frac{x^2 + 2x + 4}{x(x+4)} = \frac{x - 2}{x + 4} \]

5. Cross-multiply to solve for \( x \):

\[ (x^2 + 2x + 4)(x + 4) = (x - 2)x(x + 4) \]

Expand both sides:

\[ x^3 + 4x^2 + 2x^2 + 8x + 4x + 16 = x^3 + 4x^2 - 2x^2 - 8x \]

Combine like terms:

\[ x^3 + 6x^2 + 12x + 16 = x^3 + 2x^2 - 8x \]

Subtract \( x^3 + 2x^2 - 8x \) from both sides:

\[ x^3 + 6x^2 + 12x + 16 - (x^3 + 2x^2 - 8x) = 0 \]

\[ x^3 + 6x^2 + 12x + 16 - x^3 - 2x^2 + 8x = 0 \]

\[ 4x^2 + 20x + 16 = 0 \]

6. Solve the quadratic equation:

\[ 4x^2 + 20x + 16 = 0 \]

Divide by 4:

\[ x^2 + 5x + 4 = 0 \]

Factorize:

\[ (x + 4)(x + 1) = 0 \]

So the solutions are:

\[ x + 4 = 0 \quad \text{or} \quad x + 1 = 0 \]

\[ x = -4 \quad \text{or} \quad x = -1 \]

7. Check for extraneous solutions:

Substitute \( x = -4 \) into the original equation:

\[ \frac{-4 + 1}{-4} - \frac{3(-4)}{-4(-4+4)} = \frac{-4 - 2}{-4 + 4} \]

The right-hand side is undefined because the denominator is zero, so \( x = -4 \) is not a valid solution.

Substitute \( x = -1 \) into the original equation:

\[ \frac{-1 + 1}{-1} - \frac{3(-1)}{-1(-1+4)} = \frac{-1 - 2}{-1 + 4} \]

\[ 0 - \frac{-3}{-3} = \frac{-3}{3} \]

\[ 0 - 1 = -1 \]

So \( x = -1 \) is a valid solution.

Therefore, the solution to the equation is:

\[ x = -1 \]