The partial fraction decomposition of the rational function $\frac{1}{x^2 - 1}$ can be done by recognizing that $x^2 - 1$ is a difference of squares. Specifically:

$x^2 - 1 = (x - 1)(x + 1)$

Now, we can express $\frac{1}{x^2 - 1}$ as a sum of partial fractions:

$\frac{1}{x^2 - 1} = \frac{A}{x - 1} + \frac{B}{x + 1}$

To find $A$ and $B$, we multiply both sides by $(x - 1)(x + 1)$ to clear the denominators:

$1 = A(x + 1) + B(x - 1)$

Expanding the right-hand side:

$1 = A(x + 1) + B(x - 1) = A x + A + B x - B = (A + B)x + (A - B)$

Now, we can equate the coefficients of like terms. From the equation:

$(A + B)x + (A - B) = 1$

We compare the coefficients of $x$ and the constant terms:

1. $A + B = 0$ (coefficient of $x$)
2. $A - B = 1$ (constant term)

Solving this system of equations:

1. $A + B = 0$ implies $A = -B$
2. Substituting $A = -B$ into $A - B = 1$:

$- B - B = 1 \quad \Rightarrow \quad -2B = 1 \quad \Rightarrow \quad B = -\frac{1}{2}$

Now, since $A = -B$, we have:

$A = \frac{1}{2}$

Thus, the partial fraction decomposition of $\frac{1}{x^2 - 1}$ is:

$\frac{1}{x^2 - 1} = \frac{1/2}{x - 1} - \frac{1/2}{x + 1}$

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Related Questions:

1. What is the partial fraction decomposition of $\frac{1}{x^2 - 4}$?
2. How do you handle partial fraction decomposition when the denominator has repeated roots?
3. Can you decompose $\frac{2x+3}{x^2-1}$ into partial fractions?
4. What are the steps for decomposing rational functions with quadratic factors in the denominator?
5. How does partial fraction decomposition work for higher-degree polynomials?

Tip:

When decomposing rational functions, always factor the denominator first. This simplifies the decomposition into simpler fractions.