You need to decompose (x-9)/((x+3)(x-6)(x+4)) as a partial fraction.
You're looking for a,b,c in RR such that (x-9)/((x+3)(x-6)(x+4)) = a/(x+3) + b/(x-6) + c/(x+4). I'm gonna show you how to find a only, because b and c are to be found in the exact same way.
You multiply both sides by x+3, this will make it disappear from the denominator of the left side and make it appear next to b and c.
(x-9)/((x+3)(x-6)(x+4)) = a/(x+3) + b/(x-6) + c/(x+4) iff (x-9)/((x-6)(x+4)) = a + (b(x+3))/(x-6) + (c(x+3))/(x+4). You evaluate this at x-3 in order to make b and c disappear and find a.
x = -3 iff 12/9 = 4/3= a. You do the same for b and c, except that you multiply both sides by their respective denominators, and you will find out that b = -1/30 and c = -13/10.
It means we now have to integrate 4/3intdx/(x+3) - 1/30intdx/(x-6) - 13/10intdx/(x+4) = 4/3lnabs(x+3) -1/30lnabs(x-6) - 13/10lnabs(x+4)
