#int (dx)/sqrt(e^(2x)+12e^x+37)#
=#int (dx)/sqrt((e^x+1)^2+6^2)#
=#int (e^x*dx)/[e^x*sqrt((e^x+1)^2+6^2)]#
After using #e^x+1=6tany#, #e^x=6tany-1# and #e^x*dx=6(secu)^2*du# transforms, this integral became
#int (6(secy)^2*dy)/[(6tany-1)*6secy]#
=#int (secy*dy)/[(6tany-1)#
After using #z=tan(y/2)#, #dy=(2dz)/(z^2+1)#, #tany=(2z)/(1-z^2)# and #secy=(1+z^2)/(1-z^2)# transforms, it became
=#int [(1+z^2)/(1-z^2)(2dz)/(z^2+1)]/[6*(2z)/(1-z^2)-1]#
=#int [(2z*dz)/(1-z^2)]/[(z^2-12z-1)/(1-z^2)]#
=#int (2dz)/(z^2+12z-1)#
=#int (2dz)/[(z+6)^2-37]#
=#int (2dz)/[(z+6+sqrt37)*(z+6-sqrt37)]#
=#sqrt37/37int dz/(z+6-sqrt37)-sqrt37/37int dz/(z+6+sqrt37)#
=#sqrt37/37Ln(z+6-sqrt37)#-#sqrt37/37Ln(z+6+sqrt37)+C#
=#sqrt37/37Ln((z+6-sqrt37)/(z+6+sqrt37))+C#
=#sqrt37/37Ln((tan(y/2)+6-sqrt37)/(tan(y/2)+6+sqrt37))+C#
After using #e^x+1=6tany# and #y=arctan((e^x+1)/6)# inverse transforms, I found
#int (dx)/sqrt(e^(2x)+12e^x+37)#
=#sqrt37/37Ln((tan[1/2*arctan((e^x+1)/6)]+6-sqrt37)/(tan[1/2*arctan((e^x+1)/6)]+6+sqrt37))+C#
