How do you write the partial fraction decomposition of the rational expression # (x^3 - 2) / (x^4 - 1)#?

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Answer 1 · 2015-12-05T20:41:51

#(x^3-2)/(x^4-1) = (x+2)/(2(x^2+1))+3/(4(x+1))-1/(4(x-1))#

First, note that we can factor the denominator as

#x^4 - 1 = (x^2+1)(x^2-1)=(x^2+1)(x+1)(x-2)#

#(x^3-2)/(x^4-1) = (Ax+B)/(x^2+1)+C/(x+1)+D/(x-1)#

Multiplying through by #(x^2+1)(x+1)(x-1)# gives

#x^3-2#

#= (Ax+B)(x^2-1) + C(x^2+1)(x-1) + D(x^2+1)(x+1)#

#= (A + C + D)x^3 + (B - C + D)x^2 + (-A + C + D)x + (-B -C+D)#

Equating coefficients then gives us the system

#{(A+C+D = 1), (B-C+D = 0), (-A + C + D = 0), (-B-C+D = -2):}#

Solving this gives us
#{(A = 1/2), (B = 1), (C = 3/4), (D = -1/4):}#

Substituting back, we get the final result.

#(x^3-2)/(x^4-1) = (1/2x+1)/(x^2+1)+(3/4)/(x+1)+(-1/4)/(x-1)#

#= (x+2)/(2(x^2+1))+3/(4(x+1))-1/(4(x-1))#