The integral is
#I=int((1+lnx)^2dx)/x^2#
#=int((1+2lnx+(lnx)^2)dx)/x^2#
#=intdx/x^2+int(2lnxdx)/x^2+int((lnx)^2dx)/x^2#
#=I_1+I_2+I_3#
Therefore,
#I_1=intdx/x^2=intx^-2dx=-1/x#
For the calculation of #I_2#, perform an integration by parts
#intuv'=uv-intu'v#
#u=lnx#, #=>#, #u'=1/x#
#v'=2/x^2#, #=>#, #v=-2/x#
So,
#I_2=-2lnx/x-2int(-dx)/x^2#
#=-2lnx/x-2/x#
For the calculation of #I_3#, perform an integration by parts
#u=(lnx)^2#, #=>#, #u'=2lnx/x#
#v'=1/x^2#, #=>#, #v=-1/x#
#I_3=-(lnx)^2/x-2int(-lnxdx)/x^2#
#=-(lnx)^2/x-2lnx/x-2/x#
Finally,
#I=I_1+I_2+I_3#
#=-1/x-2lnx/x-2/x-(lnx)^2/x-2lnx/x-2/x+C#
#=-5/x-4lnx/x-(lnx)^2/x+C#
