The denominator is
#xsqrt(3+x^2)=xsqrt(3(1+(x/sqrt3)^2))#
#=sqrt3xsqrt((1+(x/sqrt3)^2))#
Perform the substitution
#x/sqrt3=tantheta#
#dx/sqrt3=sec^2theta d theta#
#sqrt((1+(x/sqrt3)^2))=sqrt(1+tan^2theta)=sqrt(sec^2theta)=sectheta#
Therefore, the integral is
#I=int(dx)/(xsqrt(3+x^2))=int(sqrt3sec^2theta d theta)/(sqrt3 tan thetasectheta)#
#=int(secthetad theta)/tan theta#
#=int(d theta)/sin theta#
#=int(csctheta d theta)#
#=int(csctheta(csctheta+cottheta)d theta)/(csctheta+cottheta)#
#=int((csc^2theta+csc thetacottheta)d theta)/(csctheta+cottheta)#
Perform the substitution
#v=csctheta+cottheta#, #=>#, #dv=(-cotthetacsctheta-csc^2theta)d theta#
Therefore,
#I=int(-dv)/(v)#
#=-ln(v)#
#=-ln(csctheta+cottheta)#
#tan theta=x/sqrt3#, #=>#, #csctheta=sqrt(3+x^2)/x#
#tan theta=x/sqrt3#, #=>#, #cottheta=sqrt(3)/x#
Finally,
#I=-ln(sqrt(3+x^2)/x+sqrt(3)/x)#
#=-ln((sqrt(3+x^2)+sqrt3)/(x))#
#=ln(|x/(sqrt(3+x^2)+sqrt3)|)+C#
