将 $\frac{x^2 + 1}{x^2 - 1}$ 分解为部分分式。

1. Extract the polynomial part since the fraction is improper

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

2. Set up the partial fraction identity for the proper fraction part

Factorise the denominator: $x^2 - 1 = (x - 1)(x + 1)$
Let $\frac{2}{(x - 1)(x + 1)} = \frac{A}{x - 1} + \frac{B}{x + 1}$
Multiply by $(x - 1)(x + 1)$:
$2 = A(x + 1) + B(x - 1)$

3. Solve for the constants A and B

Let $x = 1$:
$2 = A(1 + 1) \implies 2A = 2 \implies A = 1$

Let $x = -1$:
$2 = B(-1 - 1) \implies -2B = 2 \implies B = -1$

4. Write the final partial fraction expression

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