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,

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

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