#(x^2-2x+3)/((x+3)(x-1)(x+3))=A/(x+3)^2+B/(x+3)+C/(x-1)#
Use LCD #=(x+3)^2(x-1)#, this is the denominator
#(x^2-2x+3)/((x+3)^2(x-1))=(A(x-1)+B(x+3)(x-1)+C(x+3)^2)/((x+3)^2(x-1))#
#(x^2-2x+3)/((x+3)^2(x-1))=(Ax-A+Bx^2+2Bx-3B+Cx^2+6Cx+9C)/((x+3)^2(x-1))#
the equations are
#B+C=1#
#A+2B+6C=-2#
#-A-3B+9C=3#
Simultaneous solution results to
#A=-9/2#
#B=7/8#
#C=1/8#
#(x^2-2x+3)/((x+3)(x-1)(x+3))=A/(x+3)^2+B/(x+3)+C/(x-1)#
#(x^2-2x+3)/((x+3)(x-1)(x+3))=(-9/2)/(x+3)^2+(7/8)/(x+3)+(1/8)/(x-1)#
Let us now integrate
#int ((x^2-2x+3)dx)/((x+3)(x-1)(x+3))=#
#int (-9/2dx)/(x+3)^2+int (7/8dx)/(x+3)+int (1/8dx)/(x-1)#
#int ((x^2-2x+3)*dx)/((x+3)(x-1)(x+3))=#
#(9/2)/(x+3)+7/8*ln(x+3)+1/8*ln(x-1)+C#
God bless...I hope the explanation is useful.
