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Integrate 1/(x^2 - 1) dx using partial fractions

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Answer
12ln⁡∣x−1x+1∣+C\boxed{\frac{1}{2} \ln\left| \frac{x - 1}{x + 1} \right| + C}

Differentiating this result using the chain rule yields the original integrand 1x2−1\frac{1}{x^2 - 1}.

Problem: Evaluate the indefinite integral ∫1x2−1 dx\int \frac{1}{x^2 - 1} \, dx using partial fractions.

Steps

  1. Factor the denominator
    ∫1(x−1)(x+1) dx\int \frac{1}{(x - 1)(x + 1)} \, dx
  2. Apply partial fraction decomposition
    ∫(1/2x−1−1/2x+1)dx\int \left( \frac{1/2}{x - 1} - \frac{1/2}{x + 1} \right) dx
  3. Integrate each term
    12ln⁡∣x−1∣−12ln⁡∣x+1∣+C\frac{1}{2} \ln|x - 1| - \frac{1}{2} \ln|x + 1| + C
  4. Combine logarithms using logarithm properties
    12ln⁡∣x−1x+1∣+C\frac{1}{2} \ln\left| \frac{x - 1}{x + 1} \right| + C

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