How do you express #(x-1)/(x^3 +x)# in partial fractions?

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Answer 1 · 2016-08-14T00:06:50

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

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

#(x-1)/(x(x^2+1)) = A/x + (Bx+C)/(x^2+1)#

#x-1 = A(x^2+1) + x(Bx+C)#

#x-1 = Ax^2 + A + Bx^2 + Cx#

Comparing coefficients:

#0 = A + B#

#1 = C#

#-1 = A implies B = 1#

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