Official solution
\[{{dy} \over {dx}} - {{y + 3x} \over {\ln (y + 3x)}} + 3 = 0\]
Let \[\ln (y + 3x) = t\]
\[{1 \over {y + 3x}}.\left( {{{dy} \over {dx}} + 3} \right) = {{dt} \over {dx}}\]
\[ \Rightarrow \left( {{{dy} \over {dx}} + 3} \right) = {{y + 3x} \over {\ln (y + 3x)}}\]
\[ \therefore \] \[\left( {y + 3x} \right){{dt} \over {dx}} = {{y + 3x} \over t}\]
\[ \Rightarrow tdt = dx\]
\[{{{t^2}} \over 2} = x + c\]
\[{1 \over 2}{\left( {\ln (y + 3x)} \right)^2} = x + c\]