Answer
$$\frac{{{{\left( {3\ln x + 2} \right)}^5}}}{{15}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{{\left( {3\ln x + 2} \right)}^4}}}{x}} dx \cr
& {\text{set }}u = 3\ln x + 2{\text{ then }}\frac{{du}}{{dx}} = \frac{3}{x},\,\,\,\,\,\,\frac{1}{3}du = \frac{1}{x}dx \cr
& {\text{write the integrand in terms of }}u \cr
& \int {\frac{{{{\left( {3\ln x + 2} \right)}^4}}}{x}} dx = \int {{u^4}} \left( {\frac{1}{3}du} \right) \cr
& = \frac{1}{4}\int {{u^4}} du \cr
& {\text{integrate by using the power rule }}\int {{u^n}} du = \frac{{{u^{n + 1}}}}{{n + 1}} + C \cr
& = \frac{1}{3}\left( {\frac{{{u^5}}}{5}} \right) + C \cr
& = \frac{{{u^5}}}{{15}} + C \cr
& {\text{replace }}3\ln x + 2{\text{ for }}u \cr
& = \frac{{{{\left( {3\ln x + 2} \right)}^5}}}{{15}} + C \cr} $$