How do you integrate int (1-x^2)/((x+8)(x-5)(x+2)) ∫1−x2(x+8)(x−5)(x+2) using partial fractions?

1 Answer
Narad T.
Nov 4, 2016

The answer is =-21/26ln(x+8)-24/91ln(x-5)+1/14ln(x+2)+C=−2126ln(x+8)−2491ln(x−5)+114ln(x+2)+C

Explanation:

Let's start by the decomposition into partial fractions
(1-x^2)/((x+8)(x-5)(x+2))=A/(x+8)+B/(x-5)+C/(x+2)1−x2(x+8)(x−5)(x+2)=Ax+8+Bx−5+Cx+2
Upon simplification
1-x^2=A(x-5)(x+2)+B(x+8)(x+2)+C(x+8)(x-5)1−x2=A(x−5)(x+2)+B(x+8)(x+2)+C(x+8)(x−5)
Let x=5x=5=>⇒-24=91B−24=91B=>⇒B=-24/91B=−2491
Let x=-2x=−2=>⇒-3=-42C−3=−42C=>⇒C=1/14C=114
Let x=-8x=−8=>⇒-63=78A−63=78A=>⇒A=-21/26A=−2126
int((1-x^2)dx)/((x+8)(x-5)(x+2))=int(-21/26dx)/(x+8)+int(-24/91dx)/(x-5)+int(1/14dx)/(x+2)∫(1−x2)dx(x+8)(x−5)(x+2)=∫−2126dxx+8+∫−2491dxx−5+∫114dxx+2
=-21/26ln(x+8)-24/91ln(x-5)+1/14ln(x+2)+C=−2126ln(x+8)−2491ln(x−5)+114ln(x+2)+C

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