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

1 Answer
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#