Let complete the square in the denominator
#9x^2-6x+2=9x^2-6x+1+2-1=(3x-1)^2+1#
Perform the substitution
#u=3x-1#, #=>#, #du=3dx#
Therefore,
#int(dx)/(sqrt(9x^2-6x-2))=1/3int(du)/sqrt(u^2+1)#
Let #u=tantheta#, #=>#, #du=sec^2thetad theta#
Therefore,
#int(dx)/(sqrt(9x^2-6x-2))=1/3int(sec^2thetad theta)/(sectheta)#
#=1/3intsecthetad theta#
#=1/3int(sectheta(sectheta+tantheta)d theta)/(sectheta+tantheta)#
Let #v=sectheta+tantheta#
#dv=(sec^2theta +secthetatantheta)d theta#
So,
#int(dx)/(sqrt(9x^2-6x-2))=1/3int(dv)/v#
#=1/3lnv#
#=1/3ln(sectheta+tantheta)#
#=1/3ln(u+sqrt(u^2+1))#
#=1/3ln(|(3x-1)+sqrt((3x-1)^2+1)|)+C#