1. **State the problem:** We want to find the integral of the function $$\frac{1}{x^2 - 1}$$. 2. **Recall the formula and rules:** The integral of a rational function can often be solved by partial fraction decomposition when the denominator factors. 3. **Factor the denominator:** $$x^2 - 1 = (x - 1)(x + 1)$$. 4. **Set up partial fractions:** $$\frac{1}{x^2 - 1} = \frac{1}{(x - 1)(x + 1)} = \frac{A}{x - 1} + \frac{B}{x + 1}$$ 5. **Multiply both sides by the denominator:** $$1 = A(x + 1) + B(x - 1)$$ 6. **Expand and group terms:** $$1 = A x + A + B x - B = (A + B) x + (A - B)$$ 7. **Equate coefficients:** For the coefficient of $$x$$: $$A + B = 0$$ For the constant term: $$A - B = 1$$ 8. **Solve the system:** From $$A + B = 0$$, we get $$B = -A$$. Substitute into $$A - B = 1$$: $$A - (-A) = 1 \Rightarrow 2A = 1 \Rightarrow A = \frac{1}{2}$$ Then $$B = -\frac{1}{2}$$. 9. **Rewrite the integral:** $$\int \frac{1}{x^2 - 1} dx = \int \frac{1/2}{x - 1} dx + \int \frac{-1/2}{x + 1} dx$$ 10. **Integrate each term:** $$= \frac{1}{2} \int \frac{1}{x - 1} dx - \frac{1}{2} \int \frac{1}{x + 1} dx$$ $$= \frac{1}{2} \ln|x - 1| - \frac{1}{2} \ln|x + 1| + C$$ 11. **Combine logarithms:** $$= \frac{1}{2} \ln \left| \frac{x - 1}{x + 1} \right| + C$$ **Final answer:** $$\int \frac{1}{x^2 - 1} dx = \frac{1}{2} \ln \left| \frac{x - 1}{x + 1} \right| + C$$