To solve the equation \( \frac{1}{x - 5} = \frac{1}{x + 3} + \frac{x - 1}{x^2 - 2x - 15} \), we start by simplifying it.
The denominator \( x^2 - 2x - 15 \) can be factored: \[ x^2 - 2x - 15 = (x - 5)(x + 3) \]
So we rewrite the equation as: \[ \frac{1}{x - 5} = \frac{1}{x + 3} + \frac{x - 1}{(x - 5)(x + 3)} \]
Now we can combine the right-hand side: \[ \frac{1}{x + 3} + \frac{x - 1}{(x - 5)(x + 3)} = \frac{(x - 5) + (x - 1)}{(x - 5)(x + 3)} = \frac{2x - 6}{(x - 5)(x + 3)} = \frac{2(x - 3)}{(x - 5)(x + 3)} \]
Now we can set the two sides equal: \[ \frac{1}{x - 5} = \frac{2(x - 3)}{(x - 5)(x + 3)} \]
Next, we can cross-multiply: \[ (x + 3) = 2(x - 3) \] Expanding that gives: \[ x + 3 = 2x - 6 \]
Now, we can simplify to find \( x \): \[ 3 + 6 = 2x - x \] \[ 9 = x \]
Thus, the solution is: \[ \boxed{x = 9} \]