Answer
$$\frac{{{x^4}\ln x}}{4} - \frac{{{x^4}}}{{16}} + C$$
Work Step by Step
$$\eqalign{
& \int {{x^3}} \ln xdx \cr
& {\text{setting }}\,\,\,\,\,\,u = \ln x{\text{ then }}du = \frac{1}{x}dx\,\,\,\,\,\,\,\,\,\, \cr
& {\text{and}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dv = {x^3}dx{\text{ then }}v = \frac{{{x^4}}}{4} \cr
& {\text{Substituting these values into the formula for integration by parts}} \cr
& \int u dv = uv - \int {vdu} \cr
& \int {\ln x} \left( {{x^3}} \right)dx = \left( {\ln x} \right)\left( {\frac{{{x^4}}}{4}} \right) - \int {\frac{{{x^4}}}{4}\left( {\frac{1}{x}} \right)} dx \cr
& \int {{x^3}} \ln xdx = \frac{{{x^4}\ln x}}{4} - \int {\frac{{{x^3}}}{4}} dx \cr
& \int {{x^3}} \ln xdx = \frac{{{x^4}\ln x}}{4} - \frac{1}{4}\int {{x^3}} dx \cr
& {\text{integrate using the power rule }}\int {{x^n}} dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C \cr
& \int {{x^3}} \ln xdx = \frac{{{x^4}\ln x}}{4} - \frac{1}{4}\left( {\frac{{{x^4}}}{4}} \right) + C \cr
& {\text{simplifying}} \cr
& \int {{x^3}} \ln xdx = \frac{{{x^4}\ln x}}{4} - \frac{{{x^4}}}{{16}} + C \cr} $$