First you'll need to split up the integrand into partial fractions.
#(x-3x^2)/((x+9)(x-5)(x-7)) = A/(x+9)+B/(x-5)+C/(x-7)#
#x-3x^2 = A(x-5)(x-7) + B(x+9)(x-7) + C(x+9)(x-5)#
We'll substitute in the roots, #x=-9,5,7#, to find each of the numerator constants.
#x = -9#
#(-9)-3xx(-9)^2 = A xx (-9-5) xx (-9-7)#
#-252 = A xx -14 xx -16#
#A = 9/8#
#x = 5#
#(5)-3(5)^2 = B xx (5+9) xx (5-7)#
#-70 = B xx 14 xx -2#
#B = 5/2#
#x = 7#
#(7)-3(7)^2 = C xx (7 + 9) xx (7-2)#
#-140 = C xx 16 xx 5#
#C = -7/4#
Therefore, we now have that
#(x-3x^2)/((x+9)(x-5)(x-7)) = (9/8)/(x+9)+(5/2)/(x-5)-(7/4)/(x-7)#
Integrating #a/(x+b)# gives you #aln(x+b)#, so
#int (9/8)/(x+9)+(5/2)/(x-5)-(7/4)/(x-7)dx =#
#9/8ln(x+9) + 5/2ln(x-5) - 7/4ln(x-7) + C#
#= ln(((x+9)^(9/8)(x-5)^(5/2))/(x-7)^(7/4)) + C#
