How do you integrate f(x)=(x^2-2)/((x+4)(x-2)(x-2))f(x)=x2−2(x+4)(x−2)(x−2) using partial fractions?

1 Answer
Narad T.
Feb 16, 2017

The answer is =7/18ln(|x+4|)+11/18ln(|x-2|)-1/(3(x-2))+C=718ln(|x+4|)+1118ln(|x−2|)−13(x−2)+C

Explanation:

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)x2−2(x+4)(x−2)2=Ax+4+B(x−2)2+Cx−2

=(A(x-2)^2+B(x+4)+C(x-2)(x+4))/((x+4)(x-2)^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)x2−2=A(x−2)2+B(x+4)+C(x−2)(x+4)

Let x=-4x=−4, =>⇒, 14=36A14=36A, =>⇒, A=7/18A=718

Let x=2x=2, =>⇒, 2=6B2=6B, =>⇒, B=1/3B=13

Coefficients of x^2x2,

1=A+C1=A+C, =>⇒, C=1-A=1-7/18=11/18C=1−A=1−718=1118

Therefore,

(x^2-2)/((x+4)(x-2)^2)=(7/18)/(x+4)+(1/3)/(x-2)^2+(11/18)/(x-2)x2−2(x+4)(x−2)2=718x+4+13(x−2)2+1118x−2

int((x^2-2)dx)/((x+4)(x-2)^2)=7/18intdx/(x+4)+1/3intdx/(x-2)^2+11/18intdx/(x-2)∫(x2−2)dx(x+4)(x−2)2=718∫dxx+4+13∫dx(x−2)2+1118∫dxx−2

=7/18ln(|x+4|)+11/18ln(|x-2|)-1/(3(x-2))+C=718ln(|x+4|)+1118ln(|x−2|)−13(x−2)+C

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