#(x+1)/[(2x-4)*(x-4)(x-7)]#
=#1/2*(x+1)/[(x-2)*(x-4)(x-7)]#
I decomposed #(x+1)/[(x-2)*(x-4)(x-7)]# into basic fractions,
#(x+1)/[(x-2)*(x-4)(x-7)]#
=#A/(x-2)+B/(x-4)+C/(x-7)#
After expanding denominator,
#A(x-4)(x-7)+B(x-2)(x-7)+C(x-2)(x-4)=x+1#
Set #x=2#, #10A=3#, so #A=3/10#
Set #x=4#, #-6B=5#, so #B=-5/6#
Set #x=7#, #15C=8#, so #C=8/15#
Thus,
#int (x+1)/[(2x-4)*(x-4)(x-7)]*dx#
=#1/2int (x+1)/[(2x-4)*(x-4)(x-7)]*dx#
=#1/2*3/10int (dx)/(x-2)-1/2*5/6int (dx)/(x-4)+1/2*8/15int (dx)/(x-7)#
=#3/20int (dx)/(x-2)-5/12int (dx)/(x-4)+4/15int (dx)/(x-7)#
=#3/20Ln(Abs(x-2))-5/12(Abs(x-4))+4/15Ln(Abs(x-7))+C#