#int (6x-1)/sqrt(x^2+3x+8)*dx#
=#int (12x-2)/sqrt(4x^2+12x+32)*dx#
=#int (12x-2)/sqrt((2x+3)^2+23)*dx#
=#int (12x+18-20)/sqrt((2x+3)^2+23)*dx#
=#3/2*int (8x+12)/sqrt((2x+3)^2+23)*dx#-#int 20/sqrt((2x+3)^2+23)*dx#
#A=3/2*int (8x+12)/sqrt((2x+3)^2+23)*dx#
After using #y=4x^2+12x+32# and #dy=(8x+12)*dx# transforms, #A# becomes
#A=3/2*int dy/sqrty#
=#3sqrty#
=#3sqrt(4x^2+12x+32)#
=#6sqrt(x^2+3x+8)#
#B=int 20/sqrt((2x+3)^2+23)*dx#
=#10int 2/sqrt((2x+3)^2+23)*dx#
After using #2x+3=sqrt23*tanz# and #2dx=sqrt23*(secz)^2*dz# transforms, #B# becomes
#B=int (10sqrt23*(secz)^2)/(23secz)*dz#
=#(10sqrt23)/23int secz*dz#
=#(10sqrt23)/23int (secz*(secz+tanz)*dz)/(secz+tanz)#
=#(10sqrt23)/23ln(secz+tanz)-C1#
After using #2x+3=sqrt23*tanz#, #tanz=(2x+3)/sqrt23 and #secz=#sqrt(4x^2+12x+32)/sqrt23# inverse transforms,
#B=(10sqrt23)/23ln(sqrt(4x^2+12x+32)+2x+3)-C#
Thus,
#int (6x-1)/sqrt(x^2+3x+8)*dx#
=#A-B#
=#6sqrt(x^2+3x+8)#-#(10sqrt23)/23ln(sqrt(4x^2+12x+32)+2x+3)+C#
Note: #C=C1+Ln(sqrt23)#
