To solve the equation

\[ \frac{9}{x^2 - 4} = \frac{x + 1}{x + 2} + \frac{3}{x - 2}, \]

we first recognize that \( x^2 - 4 \) can be factored as \( (x + 2)(x - 2) \). This gives us:

\[ \frac{9}{(x + 2)(x - 2)} = \frac{x + 1}{x + 2} + \frac{3}{x - 2}. \]

Next, we find a common denominator for the right-hand side of the equation. The common denominator is \( (x + 2)(x - 2) \). We rewrite each fraction with this common denominator:

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

Now, we combine these fractions:

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

Simplify the numerator:

\[ (x + 1)(x - 2) + 3(x + 2) = (x^2 - 2x + x - 2) + (3x + 6) = x^2 - x - 2 + 3x + 6 = x^2 + 2x + 4. \]

So, the right-hand side becomes:

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

Now, we equate the left-hand side and the right-hand side:

\[ \frac{9}{(x + 2)(x - 2)} = \frac{x^2 + 2x + 4}{(x + 2)(x - 2)}. \]

Since the denominators are the same, we can equate the numerators:

\[ 9 = x^2 + 2x + 4. \]

Rearrange this equation to form a standard quadratic equation:

\[ x^2 + 2x + 4 - 9 = 0, \]
\[ x^2 + 2x - 5 = 0. \]

We solve this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 2 \), and \( c = -5 \):

\[ x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-5)}}{2 \cdot 1}, \]
\[ x = \frac{-2 \pm \sqrt{4 + 20}}{2}, \]
\[ x = \frac{-2 \pm \sqrt{24}}{2}, \]
\[ x = \frac{-2 \pm 2\sqrt{6}}{2}, \]
\[ x = -1 \pm \sqrt{6}. \]

So, the solutions are:

\[ x = -1 + \sqrt{6}, \]
\[ x = -1 - \sqrt{6}. \]

We should check if these solutions make the original equation undefined. The original equation is undefined if \( x = 2 \) or \( x = -2 \), but neither \( -1 + \sqrt{6} \) nor \( -1 - \sqrt{6} \) equals 2 or -2. Therefore, both solutions are valid.

Thus, the solutions to the equation are:

\[ x = -1 + \sqrt{6}, \]
\[ x = -1 - \sqrt{6}. \]