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

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Answer 1 · 2017-01-17T10:55:02

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

Let's do the decomposition into partial fractions

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

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

We equalise the denominators

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

Let #x=-2#, #=>#,#2=9A#, #=>#, #A=2/9#

Let #x=1#, #=>#, #2=3C#, #=>#, #C=2/3#

Coefficients of #x^2#

#1=A+B#, #=>#, #B=1-A=1-2/9=7/9#

So,

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