How do you integrate int (1-x^2)/((x-9)(x-5)(x+2))dx ∫1−x2(x−9)(x−5)(x+2)dx using partial fractions?

1 Answer
Rui D.
Mar 5, 2016

-20/11ln|x-9|+6/7ln|x-5|-3/77ln|x+2|+const.−2011ln|x−9|+67ln|x−5|−377ln|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)1−x2(x−9)(x−5)(x+2)=Ax−9+Bx−5+Cx+2

Making x=0, 1 and -1x=0,1and−1 (it's convenient to make x=1x=1 and x=-1x=−1 since then 1-x^2=01−x2=0), we get

1/90=A/-9+B/-5+C/2190=A−9+B−5+C2
0=A/-8+B/-4+C/30=A−8+B−4+C3
0=A/-10+B/-6+C/10=A−10+B−6+C1

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.

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