How do you express #1/((x²)(x² + 3x + 2)) # in partial fractions?

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Answer 1 · 2016-11-15T12:33:37

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

As #1/(x^2(x^2+3x+2))=1/(x^2(x+2)(x+1))#, let the partial fractions be given by

#1/(x^2(x+2)(x+1))hArrA/x+B/x^2+C/(x+1)+D/(x+2)#

or #1/(x^2(x+2)(x+1))hArr(Ax(x+1)(x+2)+B(x+1)(x+2)+Cx^2(x+2)+Dx^2(x+1))/(x^2(x+2)(x+1))#

Now, if #x=-1#, we have #C=1# and if #x=-2#, we have #-4D=1# or #D=-1/4# and if #x=0#, we have #2B=1# or #B=1/2#.

Now comparing the terms of #x^3#, we get #A+C+D=0# and hence
#A=-C-D=-1-(-1/4)=-3/4#

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

Answer 2 · 2016-11-15T12:51:14

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

The denominator #=x^2(x^2+3x+2)=x^2(x+2)(x+1)#

The decomposition in partial fractions is

#1/(x^2(x+2)(x+1))=A/x^2+B/x+C/(x+2)+D/(x+1)#

#=(A(x+1)(x+2)+Bx(x+1)(x+2)+Cx^2(x+1)+Dx^2(x+2))/((x^2)(x+1)(x+2))#

#1=(A(x+1)(x+2)+Bx(x+1)(x+2)+Cx^2(x+1)+Dx^2(x+2))#

Let x=0, #=>##1=2A# ; #A=1/2#

Let #x=-1#, #=># #1=D#

Let #x=-2#, #=># #1=-4C# , #C=-1/4#

Coefficient of #x^3#, #=>#, #0=B+C+D#

#B=-C-D=1/4-1=-3/4#

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