How do you express #(x+4)/((x+1)(x-2)^2)# in partial fractions?

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Answer 1 · 2018-05-08T13:18:57

The answer is #=(13/3)/(x+1)+42/(x-2)^2+(-13/3)/(x-2)#

Perform the decomposition into partial fractions

#(x+40)/((x+1)(x-2)^2)=A/(x+1)+B/(x-2)^2+C/(x-2)#

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

The denominators are the same, compare the numerators

#x+40=A(x-2)^2+B(x+1)+C(x+1)(x-2)#

Let #x=-1#, #=>#, #39=9A#, #=>#, #A=39/9=13/3#

Let #x=2#, #=>#, #42=3B#, #=>#, #B=42/3=14#

Coefficients of #x^2#

#0=A+C#, #=>#, #C=-A=-13/3#

Finally,

#(x+40)/((x+1)(x-2)^2)=(13/3)/(x+1)+42/(x-2)^2+(-13/3)/(x-2)#