[Cambridge/9794/Specimen Paper for 2010] Find the value of $\int_{0}^{2} \frac{x - 1}{x^2 - 2x - 3} \, dx$.

1. Express the numerator as a scalar multiple of the derivative of the denominator.

Let $f(x) = x^2 - 2x - 3$.
Then $f'(x) = 2x - 2 = 2(x - 1)$.

Therefore, we can rewrite the integral as:
$\int_{0}^{2} \frac{x - 1}{x^2 - 2x - 3} \, dx = \frac{1}{2} \int_{0}^{2} \frac{2x - 2}{x^2 - 2x - 3} \, dx$

2. Integrate the function to obtain a natural logarithm, ensuring the use of absolute value symbols.

$= \left[ \frac{1}{2} \ln |x^2 - 2x - 3| \right]_{0}^{2}$

3. Evaluate the definite integral by substituting the limits.

$= \frac{1}{2} \ln |2^2 - 2(2) - 3| - \frac{1}{2} \ln |0^2 - 2(0) - 3|$

$= \frac{1}{2} \ln |-3| - \frac{1}{2} \ln |-3|$

$= \frac{1}{2} \ln 3 - \frac{1}{2} \ln 3$

$= 0$