Question

How do you integrate `int (1-x^2)/((x-9)(x-5)(x+2))dx` using partial fractions?

Answer 1

`-20/11ln|x-9|+6/7ln|x-5|-3/77ln|x+2|+const.`

Explanation:

We can rewrite the integrand expression in this way
`(1-x^2)/((x-9)(x-5)(x+2))=A/(x-9)+B/(x-5)+C/(x+2)`

Making `x=0, 1 and -1` (it's convenient to make `x=1` and `x=-1` since then `1-x^2=0`), we get

`1/90=A/-9+B/-5+C/2`
`0=A/-8+B/-4+C/3`
`0=A/-10+B/-6+C/1`

Or
`[[-1/9,-1/5,1/2],[-1/8,-1/4,1/3],[-1/10,-1/6,1]][[A],[B],[C]]=[[1/90],[0],[0]]`

Solving the system of variables
`Delta =[[-1/9,-1/5,1/2],[-1/8,-1/4,1/3],[-1/10,-1/6,1]]=1/36+1/150+1/96-(1/80+1/162+1/40)=(1800+432+675-810-400-1620)/64800=77/64800`
`Delta A=[[1/90,-1/5,1/2],[0,-1/4,1/3],[0,-1/6,1]]=-1/360-(-1/1620)=(-9+2)/3240=-7/3240`
`Delta B=[[-1/9,1/90,1/2],[-1/8,0,1/3],[-1/10,0,1]]=-1/2700-(-1/720)=(-4+15)/10800=11/10800`
`Delta C=[[-1/9,-1/5,1/90],[-1/8,-1/4,0],[-1/10,-1/6,0]]=1/4320-(1/3600)=(5-6)/21600=-1/21600`

So

`A=(Delta A)/Delta=(-7/3240)/(77/64800)=-1/11*20=-20/11`
`B=(Delta B)/Delta=(11/10800)/(77/64800)=1/7*6=6/7`
`C=(Delta C)/Delta=(-1/21600)/(77/64800)=-1/77*3=-3/77`

Then the original expression becomes

`=-20/11int dx/(x-9)+6/7int dx/(x-5)-3/77int dx/(x+2)`
`=-20/11ln|x-9|+6/7ln|x-5|-3/77ln|x+2|+const.`