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You tried to match \(x^2 - 2\) directly to \(A(x+1) + B(x-1)\) even though the fraction is “improper” (the numerator and denominator have the same degree), so you actually need to do polynomial long division first. If you divide \(x^2 - 2\) by \(x^2 - 1\), you get \(1 - \frac{1}{x^2-1}\). Then \(-\frac{1}{x^2-1} = \frac{-1}{(x-1)(x+1)}\) can be split correctly into partial fractions: \(-\frac{1}{(x-1)(x+1)} = -\frac{1}{2}\cdot\frac{1}{x-1} + \frac{1}{2}\cdot\frac{1}{x+1}\). Thus, \(\frac{x^2-2}{x^2-1} = 1 - \frac{1}{2}\cdot\frac{1}{x-1} + \frac{1}{2}\cdot\frac{1}{x+1}\).
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Thu, 13 Feb 2025 19:47 GMT