How do you write the partial fraction decomposition of the rational expression # (x^2+x)/((x+2)(x-1)^2)#?

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Answer 1 · 2016-11-06T06:57:15

#(x^2+x)/((x+2)(x-1)^2)=2/(9(x+2))+7/(9(x-1))+2/(3(x-1)^2)#

As the degree of numerator is less than that of denominator, we can proceed to write partial fractions as

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

or #x^2+x=A(x-1)^2+B(x-1)(x+2)+C(x+2)#............(1)

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

or #x^2+x=x^2(A+B)+x(B-2A+C)+(A-2B+2C)#............(2)

Putting #x=1# and #x=-2# in (1), we get #3C=2# and #9A=2# i.e. #C=2/3# and #A=2/9#.

Comparing terms of #x^2# in (2), #A+B=1# i.e. #B=7/9#

Hence #(x^2+x)/((x+2)(x-1)^2)=2/(9(x+2))+7/(9(x-1))+2/(3(x-1)^2)#