How do you express # (x^3 + 2) / (x^4 -16)# in partial fractions?

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Answer 1 · 2016-11-20T07:20:49

The answer is #=(3/16)/(x+2)+(5/16)/(x-2)+(x/2-1/4)/(x^2+4)#

Let's factorise the denominator

#x^4-16=(x^2-4)(x^2+4)=(x+2)(x-2)(x^2+4)#

Therefore,

#(x^3+2)/(x^4-16)=(x^3+2)/((x+2)(x-2)(x^2+4))#

The decomposition in partial fractions is

#(x^3+2)/(x^4-16)=A/(x+2)+B/(x-2)+(Cx+D)/(x^2+4)#

#(A(x-2)(x^2+4)+B(x+2)(x^2+4)+(Cx+D)(x-2)(x+2))/((x-2)(x+2)(x^2+4))#

Therefore,

#(x^3+2)=(A(x-2)(x^2+4)+B(x+2)(x^2+4)+(Cx+D)(x-2)(x+2))#

let #x=2#
#10=32B# ; #=># #B=5/16#

Let #x=-2#
#-6=-32A# ; #=>##A=3/16#

Coefficients of #x^3#
#1=A+B+C# , #=># #C=1-5/16-3/16=8/16=1/2#

Let #x=0#
#2=-8A+8B-4D# : #=># #4D=-2-3/2+5/2=-1#
#D=-1/4#

#(x^3+2)/(x^4-16)=(3/16)/(x+2)+(5/16)/(x-2)+(x/2-1/4)/(x^2+4)#