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

1 Answer
Narad T.
May 8, 2018

The answer is =(13/3)/(x+1)+42/(x-2)^2+(-13/3)/(x-2)=133x+1+42(x−2)2+−133x−2

Explanation:

Perform the decomposition into partial fractions

(x+40)/((x+1)(x-2)^2)=A/(x+1)+B/(x-2)^2+C/(x-2)x+40(x+1)(x−2)2=Ax+1+B(x−2)2+Cx−2

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

Let x=-1x=−1, =>⇒, 39=9A39=9A, =>⇒, A=39/9=13/3A=399=133

Let x=2x=2, =>⇒, 42=3B42=3B, =>⇒, B=42/3=14B=423=14

Coefficients of x^2x2

0=A+C0=A+C, =>⇒, C=-A=-13/3C=−A=−133

Finally,

(x+40)/((x+1)(x-2)^2)=(13/3)/(x+1)+42/(x-2)^2+(-13/3)/(x-2)x+40(x+1)(x−2)2=133x+1+42(x−2)2+−133x−2

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