#f(x)=int (x^2+x)/[(x+3)*(x+4)]*dx#
=#int (x^2+7x+12-6x-12)/[(x+3)*(x+4)]*dx#
=#int dx-int (6x+12)/[(x+3)*(x+4)]*dx#
=#x-int (6x+12)/[(x+3)*(x+4)]*dx+C#
Now I solved #int (6x+12)/[(x+3)*(x+4)]*dx# integral,
#(6x+12)/[(x+3)*(x+4)]=A/(x+3)+B/(x+4)#
After expanding denominator,
#A*(x+4)+B*(x+3)=6x+12#
Set #x=-4#, #-B=12#, so #B=12#
Set #x=-3#, #A=-6#
Hence,
#f(x)=int (x^2+x)/[(x+3)*(x+4)]*dx#
=#x-[int -6/(x+3)*dx+12/(x+4)*dx]+C#
=#x-(-6Ln(x+3)+12Ln(x+4))+C#
=#x+6Ln(x+3)-12Ln(x+4)+C#
After using #f(-2)=2# condition,
#-2+6Ln(-2+3)-12Ln(-2+4)+C=2#
#C+6Ln(1)-12Ln(2)-2=2#
#C+0-12Ln(2)-2=2#
#C=4+12Ln2#
Thus,
#f(x)=x+4+12Ln(2)+6Ln(x+3)-12Ln(x+4)#
