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)#
