Step 1: Analyzing the function for valid values of \( x \).
We need to check the conditions for the logarithmic and exponential parts:
- The argument of the logarithm \( x - 2 \) must be greater than 0:
\( x - 2 > 0 \Rightarrow x > 2 \) - The argument of the logarithm \( x + 1 \) must be greater than 0:
\( x + 1 > 0 \Rightarrow x > -1 \) - For the denominator \( \frac{1}{x+1} \) to be valid, it must not be 0, so:
\( x + 1 \neq 0 \Rightarrow x \neq -1 \) and \( x > 0 \) - The denominator \( x^2 - 2x - 3 \) must not be 0, so:
\( (x - 3)(x + 1) \neq 0 \Rightarrow x \neq -1, 3 \)
Step 2: Combining the conditions.
From the above conditions, we get the domain of the function:
\( x > 2 \) and \( x \neq 3 \)
Thus, the domain of the function is \( (2, \infty) - \{3\} \) .
Step 3: Conclusion.
The correct answer is option (4) \( (2, \infty) - \{3\} \).