To integrate, we should first convert #(x-9)/((x+3)(x-7)(x+4))# in to partial fractions. Let
#(x-9)/((x+3)(x-7)(x+4))hArrA/(x+3)+B/(x-7)+C/(x+4)#. Simplifying RHS
=#[A(x-7)(x+4)+B(x+3)(x+4)+C(x+3)(x-7)]/((x+3)(x-7)(x+4))# or
=#[A(x^2-3x-28)+B(x^2+7x+12)+C(x^2-4x-21)]/((x+3)(x-7)(x+4))#
=#[(A+B+C)x^2+(-3A+7B-4C)x+(-28A+12B-21C)]/((x+3)(x-7)(x+4))#
Hence #A+B+C=0#, #-3A+7B-4C=1# and #-28A+12B-21C=-9#
Solving these will give us #A=132/110#, #B=-2/110# and #C=-13/11#
Hence #(x-9)/((x+3)(x-7)(x+4))hArr132/(110(x+3))-2/(110(x-7))-13/(11(x+4))#
Hence #int(x-9)/((x+3)(x-7)(x+4))dx# =
#int[132/(110(x+3))-2/(110(x-7))-13/(11(x+4))]dx#
or
Now one can use the identity #int(1/(ax+b))dx=1/aln(ax+b)#
Hence, #int[132/(110(x+3))-2/(110(x-7))-13/(11(x+4))]dx#
= #132/110ln(x+3)-2/110ln(x-7)-13/11ln(x+4)+c#
