## Calculus with Applications (10th Edition)

$$\frac{{3{e^4} + 1}}{{16}}$$
\eqalign{ & \int_1^e {{x^3}\ln x} dx \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 & \cr & {\text{then}} \cr & \int_1^e {{x^3}\ln x} dx = \left[ {\frac{{{x^4}\ln x}}{4} - \frac{{{x^4}}}{{16}}} \right]_1^e \cr & {\text{evaluating the limits}} \cr & = \left[ {\frac{{{{\left( e \right)}^4}\ln e}}{4} - \frac{{{{\left( e \right)}^4}}}{{16}}} \right] - \left[ {\frac{{{{\left( 1 \right)}^4}\ln 1}}{4} - \frac{{{{\left( 1 \right)}^4}}}{{16}}} \right] \cr & {\text{simplifying}} \cr & = \frac{{{e^4}}}{4} - \frac{{{e^4}}}{{16}} + \frac{1}{{16}} \cr & = \frac{{3{e^4} + 1}}{{16}} \cr}