How do you write the partial fraction decomposition of the rational expression # (x^4+6)/(x^5+7x^3)#?

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Answer 1 · 2016-11-14T07:35:55

The answer is #=(-6/49)/x+(6/7)/x^3+(55/49x)/(x^2+7)#

#(x^4+6)/(x^5+7x^3)=(x^4+6)/((x^3)(x^2+7))#

Let's do the decompositiom in partial fractions

#(x^4+6)/((x^3)(x^2+7))=A/x+B/x^2+C/x^3+(Dx+E)/(x^2+7)#

#=(Ax^2(x^2+7)+Bx(x^2+7)+C(x^2+7)+x^3(Dx+E))/((x^3)(x^2+7))#

Therefore,

#x^4+6=Ax^2(x^2+7)+Bx(x^2+7)+C(x^2+7)+x^3(Dx+E)#

Let #x=0##=># #6=7C# #=># #C=6/7#

Coefficients of #x^2#, #=># #0=7A+C# ; #A=-6/49#

coefficients of #x^4# ; #=># ; #1=A+D# ; #D=1+6/49=55/49#

Coefficients of #x^3# ; #=># ; #0=B+E#

Coefficients of #x# ; #=># ; #0=7B# #=># #B=0#

#E=0#

#(x^4+6)/((x^3)(x^2+7))=(-6/49)/x+0/x^2+(6/7)/x^3+(55/49x+0)/(x^2+7)#