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## Homework Statement

Find the partial fraction decomposition of ##\displaystyle \frac{1}{x^4 + 2x^2 \cosh (2 \alpha) + 1}##

## Homework Equations

## The Attempt at a Solution

Using the identity ##\displaystyle \cosh (2 \alpha) = \frac{e^{2 \alpha} + e^{- 2\alpha}}{2}##, we can get the fraction to the form ##\displaystyle \frac{1}{(x^2 + e^{2 \alpha})(x^2 + e^{- 2 \alpha})}##. Since we have two irreducible quadratic factors, it would seem that we would now try to find A, B C, and D such that ##\displaystyle \frac{1}{(x^2 + e^{2 \alpha})(x^2 + e^{- 2 \alpha})} = \frac{Ax + B}{x^2 + e^{2 \alpha}} + \frac{Cx + D}{x^2 + e^{-2 \alpha}} ##. But in the solutions to my book, it just directly says to find A and B such that ##\displaystyle \frac{1}{(x^2 + e^{2 \alpha})(x^2 + e^{- 2 \alpha})} = \frac{A}{x^2 + e^{2 \alpha}} + \frac{B}{x^2 + e^{-2 \alpha}} ##. Why am I wrong in assuming that the numerators must be linear factors?