How do I decompose the rational expression #(x-3)/(x^3+3x)# into partial fractions?

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Answer 1 · 2018-06-17T12:39:56

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

Perform the decomposition into partial fractions after factorising the denominator

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

#=A/x+(Bx+C)/(x^2+3)#

#=(A(x^2+3)+x(Bx+C))/(x(x^2+3))#

The denominators are the same, compare the numerators

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

Let #x=0#, #=>#, #-3=3A#, #=>#, #A=-1#

Coefficients of #x^2#

#0=A+B#, #=>#, #B=-A=1#

Coefficients of #x#

#1=C#

Therefore,

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