How do you write the partial fraction decomposition of the rational expression #(x² + 2x-7) /( (x+3)(x-1)²)#?

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Answer 1 · 2017-02-28T12:05:04

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

Let 's perform the decomposition into partial fractions

#(x^2+2x-7)/((x+3)(x-1)^2)=A/(x+3)+B/(x-1)^2+C/(x-1)#

#=(A(x-1)^2+B(x+3)+C(x+3)(x-1))/((x+3)(x-1)^2)#

As the denominators are the same, we compare the numerators

#x^2+2x-7=A(x-1)^2+B(x+3)+C(x+3)(x-1)#

Let #x=-3#, #=>#, #-4=16A#, #=>#, #A=-1/4#

Let #x=1#, #=>#, #-4=4B#, #=>#, #B=-1#

Coefficients of #x^2#

#1=A+C#, #=>#, #C=1-A=1+1/4=5/4#

Therefore,

#(x^2+2x-7)/((x+3)(x-1)^2)=(-1/4)/(x+3)-1/(x-1)^2+(5/4)/(x-1)#