#int Ln(x^2+13x+40)*dx#
=#xLn(x^2+13x+40)-int x*((2x+13)*dx)/(x^2+13x+40)#
=#xLn(x^2+13x+40)-int ((2x^2+13x)*dx)/(x^2+13x+40)#
=#xLn(x^2+13x+40)-int ((2x^2+26x+80-13x-80)*dx)/(x^2+13x+40)#
=#xLn(x^2+13x+40)-int 2dx+int ((13x+80)*dx)/(x^2+13x+40)#
=#xLn(x^2+13x+40)-2x+C1+int ((52x+320)*dx)/(4x^2+52x+160)#
=#xLn(x^2+13x+40)-2x+C1+int ((26x+160)*2dx)/((2x+13)^2-3^2)#
#A=int ((26x+160)*2dx)/((2x+13)^2-3^2)#
After using #2x+13=3secy#, #2dx=3secy*tany*dy# and #x=(3secy-13)/2# transforms, #A# became
#A=int ((26*(3secy-13)/2+160)*3secy*tany*dy)/(9(tany)^2)#
=#int ((39secy-9)*secy*dy)/(3tany)#
=#13int((secy)^2*dy)/(tany)-3int (secy*dy)/tany#
=#13Ln(tany)-3int cscy*dy#
=#13/2Ln((tany)^2)-3int (cscy*(cscy+coty)*dy)/(cscy+coty)#
=#13/2Ln((secy)^2-1)+3Ln(cscy+coty)#
=#13/2Ln((secy)^2-1)+3Ln((secy+1)/tany)#
=#13/2Ln((secy)^2-1)+3Ln((secy+1)/sqrt((secy)^2-1))#
=#13/2Ln((secy)^2-1)+3/2Ln((secy+1)^2/((secy)^2-1))#
=#13/2Ln((secy)^2-1)+3/2Ln((secy+1)/(secy-1))#
After using #2x+13=3secy# and #secy=(2x+13)/3# inverse transforms, I found
#A=13/2Ln((2x+13)/3)^2-1)+3/2Ln(((2x+13)/3+1)/((2x+13)/3-1))#
=#13/2Ln((4x^2+52x+160)/9)+3/2Ln((x+8)/(x+5))#
Thus,
#int Ln(x^2+13x+40)*dx#
=#xLn(x^2+13x+40)-2x+C1+13/2Ln((4x^2+52x+160)/9)+3/2Ln((x+8)/(x+5))#
=#xLn(x^2+13x+40)-2x+13/2Ln((x^2+13x+40)#+#3/2Ln((x+8)/(x+5))+C#
=#(2x+13)/2*Ln(x^2+13x+40)-2x+3/2Ln((x+8)/(x+5))+C#
Note: #C=C1-13/2*Ln(4/9)=C1+13Ln(3/2)#
