I'm assuming that you have: #lnx^2+x#.
Then we have:
#intlnx^2+x \ dx#
#=intlnx^2 \ dx+intx \ dx#
Using the fact that #log(a^b)=bloga# and evaluating the second integral, we get:
#=int2lnx \ dx+1/2x^2+c_1#
#=2intlnx \ dx+1/2x^2+c_1#
Let's evaluate #intlnx \ dx#. Use integration by parts:
#intu \ dv=uv-intv \ du#
Let #u=lnx,dv=1#.
#:.v=x,du=1/x#
Putting together, we get:
#intlnx \ dx=xlnx-intx*1/x \ dx#
#=xlnx-int1 \ dx#
#=xlnx-x#
#=xlnx-x+c_2#
So, we get:
#=2(xlnx-x)+c_2+1/2x^2+c_1# (notice how I didn't put #2c_2# because constant ALWAYS goes after full integration)
#=2xlnx-2x+c_2+1/2x^2+c_1#
#=2xlnx-2x+1/2x^2+C#
