You can proceed by parts , that is, using the formula
#\int u dv=uv-\int v du#. See integration by parts
This means that you need to pick what you will interpret as #u# (to be differentiated) and what you will interpret as #dv# (to be integrated).
Since you do not know (or, I should say, are usually not expected to know but rather to know how to obtain: by parts) an antiderivative of #ln x# but you know its derivative, you have no choice: you differentiate #ln x# and integrate #1/sqrt x=x^(-1/2)#.
In other words,
#u=ln x# so that #du=dx/x# and #dv=x^-(1/2)# so that #v=\int x^(-1/2)dx=2 x^(1/2)#. Applying the formula above
#\int \ln x/sqrt x dx=2sqrt x ln x-2\int x^(1/2)/x dx=2sqrt x ln x-2\int x^(-1/2) dx # so that
#\int \ln x/sqrt x dx=2sqrt x ln x-4sqrt x+C#
