To solve the equation
\[ \frac{1}{x - 5} = \frac{1}{x + 3} + \frac{x - 1}{x^2 - 2x - 15}, \]
we first factor the quadratic in the denominator of the third term:
\[ x^2 - 2x - 15 = (x - 5)(x + 3). \]
Thus, we can rewrite the equation as:
\[ \frac{1}{x - 5} = \frac{1}{x + 3} + \frac{x - 1}{(x - 5)(x + 3)}. \]
Next, let's combine the right-hand side into a single fraction. The common denominator on the right side is \((x - 5)(x + 3)\):
\[ \frac{1}{x + 3} = \frac{x - 5}{(x - 5)(x + 3)}, \]
therefore,
\[ \frac{1}{x + 3} + \frac{x - 1}{(x - 5)(x + 3)} = \frac{x - 5 + (x - 1)}{(x - 5)(x + 3)} = \frac{2x - 6}{(x - 5)(x + 3)}. \]
Now we have:
\[ \frac{1}{x - 5} = \frac{2x - 6}{(x - 5)(x + 3)}. \]
Next, we can cross-multiply to eliminate the denominators:
\[ 1 \cdot (x - 5)(x + 3) = (2x - 6)(x - 5). \]
Expanding both sides:
Left-hand side:
\[ (x - 5)(x + 3) = x^2 + 3x - 5x - 15 = x^2 - 2x - 15. \]
Right-hand side:
\[ (2x - 6)(x - 5) = 2x^2 - 10x - 6x + 30 = 2x^2 - 16x + 30. \]
Setting both sides equal gives:
\[ x^2 - 2x - 15 = 2x^2 - 16x + 30. \]
Now, let's rearrange everything to one side of the equation:
\[ 0 = 2x^2 - 16x + 30 - x^2 + 2x + 15. \]
Combining like terms:
\[ 0 = x^2 - 14x + 45. \]
Now we will solve the quadratic equation \(x^2 - 14x + 45 = 0\). We can factor it as:
\[ (x - 9)(x - 5) = 0. \]
Thus, the solutions are:
\[ x - 9 = 0 \quad \Rightarrow \quad x = 9, \] \[ x - 5 = 0 \quad \Rightarrow \quad x = 5. \]
However, we need to check for extraneous solutions. Substituting \(x = 5\) back into the original equation shows that it results in a division by zero, which is undefined. Therefore, \(x = 5\) is not a valid solution.
Only valid solution:
\[ \boxed{9}. \]