How do you integrate $f(x)=(x^2-2)/((x+4)(x-2)(x-2))$ using partial fractions?

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Answer 1 · 2017-02-16T11:20:17

The answer is $=7/18ln(|x+4|)+11/18ln(|x-2|)-1/(3(x-2))+C$

Let's perform the decomposition into partial fractions

$(x^2-2)/((x+4)(x-2)^2)=A/(x+4)+B/(x-2)^2+C/(x-2)$

$=(A(x-2)^2+B(x+4)+C(x-2)(x+4))/((x+4)(x-2)^2)$

The denominators are the same, we compare the numerators

$x^2-2=A(x-2)^2+B(x+4)+C(x-2)(x+4)$

Let $x=-4$ , $=>$ , $14=36A$ , $=>$ , $A=7/18$

Let $x=2$ , $=>$ , $2=6B$ , $=>$ , $B=1/3$

Coefficients of $x^2$ ,

$1=A+C$ , $=>$ , $C=1-A=1-7/18=11/18$

Therefore,

$(x^2-2)/((x+4)(x-2)^2)=(7/18)/(x+4)+(1/3)/(x-2)^2+(11/18)/(x-2)$

$int((x^2-2)dx)/((x+4)(x-2)^2)=7/18intdx/(x+4)+1/3intdx/(x-2)^2+11/18intdx/(x-2)$

$=7/18ln(|x+4|)+11/18ln(|x-2|)-1/(3(x-2))+C$

How do you integrate f(x)=(x^2-2)/((x+4)(x-2)(x-2))f(x)=(x^2-2)/((x+4)(x-2)(x-2)) using partial fractions?