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To determine where the given function \( f(x) = \frac{\ln(x) + \tan^{-1}(x)}{x^2 - 1} \) is continuous, we need to analyze both the numerator and the denominator of the function.

### Step-by-step analysis:

1. **Numerator**: 
   The numerator is \( \ln(x) + \tan^{-1}(x) \).
   - The natural logarithm function \( \ln(x) \) is only defined for \( x > 0 \), which means that for \( \ln(x) \) to be valid, \( x \) must be positive.
   - The inverse tangent function \( \tan^{-1}(x) \) is defined for all real numbers, so it does not impose any additional restrictions.

   Therefore, the numerator is defined for \( x > 0 \).

2. **Denominator**:
   The denominator is \( x^2 - 1 \), which can be factored as:
   \[
   x^2 - 1 = (x - 1)(x + 1)
   \]
   The denominator equals zero when \( x = 1 \) or \( x = -1 \). The function is not defined where the denominator is zero, meaning \( x = 1 \) and \( x = -1 \) are points of discontinuity.

### Conclusion:
- The function \( f(x) \) is continuous where both the numerator and denominator are defined. This implies:
  - \( x > 0 \) (from the logarithm function)
  - \( x \neq 1 \) and \( x \neq -1 \) (from the denominator)

Thus, \( f(x) \) is continuous for \( 0 < x < 1 \) and \( x > 1 \), i.e., in the intervals \( (0, 1) \cup (1, \infty) \).

Would you like further details or clarifications?

Here are some related questions you might find helpful:
1. How can we analyze the differentiability of \( f(x) \)?
2. What happens to the behavior of \( f(x) \) as \( x \) approaches 1 or -1?
3. How do the inverse trigonometric functions impact the continuity of other functions?
4. Can you describe the domain of the function \( f(x) \) if the numerator were \( e^x \) instead of \( \ln(x) \)?
5. What is the significance of the function being undefined at \( x = 1 \) and \( x = -1 \)?

**Tip**: When checking continuity, always ensure that both the numerator and denominator are defined and examine points where the denominator could become zero.

Where is the function continuous? f(x) = (ln(x) + tan⁻¹(x)) / (x² - 1)

Explore where the function f(x) = (ln(x) + tan⁻¹(x)) / (x² - 1) is continuous. This solution provides a detailed breakdown of the numerator and denominator, factoring and restrictions imposed by logarithmic and inverse trigonometric functions, with a focus on continuity. Suitable for college-level calculus students.

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