To integrate $\int \frac{2}{x^2 - 1} \, dx$, we can proceed as follows:

1. Rewrite the integrand by factoring the denominator: $\int \frac{2}{x^2 - 1} \, dx = \int \frac{2}{(x - 1)(x + 1)} \, dx$

2. Use partial fraction decomposition. We can write: $\frac{2}{(x - 1)(x + 1)} = \frac{A}{x - 1} + \frac{B}{x + 1}$ Multiplying both sides by $(x - 1)(x + 1)$ gives: $2 = A(x + 1) + B(x - 1)$ Expanding and combining like terms, we have: $2 = (A + B)x + (A - B)$ Equating coefficients, we get the system: $A + B = 0 \quad \text{and} \quad A - B = 2$ Solving this system, we find $A = 1$ and $B = -1$.

3. Rewrite the integrand using the partial fractions: $\int \frac{2}{(x - 1)(x + 1)} \, dx = \int \left( \frac{1}{x - 1} - \frac{1}{x + 1} \right) \, dx$

4. Integrate term by term: $= \int \frac{1}{x - 1} \, dx - \int \frac{1}{x + 1} \, dx$ The integral of $\frac{1}{x - 1}$ is $\ln |x - 1|$, and the integral of $\frac{1}{x + 1}$ is $\ln |x + 1|$. Thus, $= \ln |x - 1| - \ln |x + 1| + C$

5. Combine the logarithms: $= \ln \left| \frac{x - 1}{x + 1} \right| + C$

Final Answer:

$\int \frac{2}{x^2 - 1} \, dx = \ln \left| \frac{x - 1}{x + 1} \right| + C$

Would you like more details on any step or have any questions?

Here are some related questions:

1. How does partial fraction decomposition work in integration?
2. How can logarithmic properties simplify integration results?
3. What if the integrand were $\frac{2}{x^2 + 1}$ instead?
4. How do we handle integrals with irreducible quadratic denominators?
5. How is the constant of integration determined in indefinite integrals?

Tip: Always check if an integrand can be simplified or rewritten before starting integration.