Question

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

Answer 1

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`