Given: #int(3x-2x^2)/((x+9)(x+2)(x-5))dx# using partial fractions
Partial Fractions:
#(3x-2x^2)/((x+9)(x+2)(x-5)) = A/(x+9) + B/(x+2) + C/(x-5)#
#3x-2x^2= A(x+2)(x-5) + B(x+9)(x-5) + C(x+9)(x+2)#
#= A(x^2-3x-10)+B(x^2+4x-45)+C(x^2+11x+18)#
rearrange:
#3x-2x^2= x^2(A+B+C) +x(-3A+4B+11C) + (-10A-45B+18C)#
Write 3 equations:
(1):#" "A+B+C = -2#
(2):#" "-3A+4B+11C = 3#
(3):#" "-10A-45B+18C = 0#
Combine equations to solve:
3*(1)+(2): #3A + 3B + 3C = -6#
#" "ul(-3A+4B+11C = 3" ")#
#" "7B +14C = -3#
10(2) - 3(3): #-30A +40B+110C =30#
#" "ul(30A +135B-54C = 0" ")#
#" "175B+56C = 30#
Combine two new equations to solve for B:
#175B+56C = 30#
#ul(-28B-56C=12)#
#147B" "=42; " " B = 2/7 = 4/14#
Back-substitute to find C:
#7*2/7 +14C = -3; " "14C = -5; " " C = -5/14#
Back-substitute to find A:
#A +2/7 -5/14 =-2; " " A = -27/14#
#int(3x-2x^2)/((x+9)(x+2)(x-5))dx = int(-27 dx)/(14(x+9)) +int(2 dx)/(7(x+2)) +int(-5 dx)/(14(x-5)) #
#= 1/14(4ln(x+2)-27ln(x+9)-5ln(x-5))+C#
