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Chapter 5: Problem 28

Write the partial fraction decomposition of each rational expression. $$\frac{x^{2}}{(x-1)^{2}(x+1)^{2}}$$

Short Answer

Expert verified
\(\frac{x^2}{(x-1)^2(x+1)^2} = \frac{-2}{x-1} + \frac{1}{(x-1)^2} + \frac{1}{x+1} - \frac{1}{(x+1)^2}\

Step by step solution

01

Write down the partial fractions' general form

Write down the general form of the partial fractions according to the factorization of the denominator: \[\frac{x^2}{(x-1)^2(x+1)^2} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+1} + \frac{D}{(x+1)^2}\]
02

Multiply through by the denominator of the left side

To eliminate the denominator on the right, multiply every term by \((x-1)^2(x+1)^2\). This leads to: \(x^2 = A(x-1)(x+1)^2 + B(x+1)^2 + C(x-1)^2(x+1) + D(x-1)^2 \).
03

Determine the values of A, B, C, D

This is a polynomial identity and it must hold for every x. Choose smart values to isolate A, B, C, and D. For A and B, use x=1 which will nullify all terms containing \((x-1)\). For C and D, use x=-1 which will cancel out all terms with \((x+1)\). After plugging these values, one gets: For A and B, \(1 = B\); for C and D, \(1 = -D\). This gives B=1, D=-1. For getting A and C, use 0 because it minimizes the expression: \(0 = A + B + C + D\). Plugging known values: -1=A+C, hence A=-2, C=1.
04

Write down the solution

Plug A, B, C, D into the general form of the partial fractions: \[\frac{x^2}{(x-1)^2(x+1)^2} = \frac{-2}{x-1} + \frac{1}{(x-1)^2} + \frac{1}{x+1} - \frac{1}{(x+1)^2}\]

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