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

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.#