Hello,
Answer : #x ln(3x) - x + cte#.
Remark that
#d/dx (X ln(X) - X) = 1.ln(X) + X/X - 1 = ln(X)#,
therefore
#int ln(X) dX = X ln(X) - X + cte#
Now, it's easy because, #ln(3x) = ln(3) + ln(x)#, therefore,
#int ln(3x) dx = ln(3)x + x ln(x) - x + cte = x ln(3x) - x + cte#
PS. There is a classical trick to find #int ln(x) dx# if you don't have any idea to test #x ln(x) -x# : an integration by parts. The general formula is
#int u'v = uv - int uv'# (because #(uv)' = u'v + uv'#).
Now, write
#int 1 \cdot ln(x) dx = \int u' v#, with #u=x# and #v=ln(x)#. You get
#int ln(x) dx = x ln(x) - \int x/x dx = x ln(x) -x + cte#
