How do you integrate int (1-x^2)/((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

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)
Upon simplification
1-x^2=A(x-5)(x+2)+B(x+8)(x+2)+C(x+8)(x-5)
Let x=5=>-24=91B=>B=-24/91
Let x=-2=>-3=-42C=>C=1/14
Let x=-8=>-63=78A=>A=-21/26
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)
=-21/26ln(x+8)-24/91ln(x-5)+1/14ln(x+2)+C

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