## Calculus with Applications (10th Edition)

$$\frac{{{x^2}\ln x}}{2} - \frac{{{x^2}}}{4} + C$$
\eqalign{ & \int x \ln xdx \cr & {\text{setting }}\,\,\,\,\,\,u = \ln x{\text{ then }}du = \frac{1}{x}dx\,\,\,\,\,\,\,\,\,\, \cr & {\text{and}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dv = xdx{\text{ then }}v = \frac{{{x^2}}}{2} \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 \right)dx = \left( {\ln x} \right)\left( {\frac{{{x^2}}}{2}} \right) - \int {\frac{{{x^2}}}{2}\left( {\frac{1}{x}} \right)} dx \cr & \int x \ln xdx = \frac{{{x^2}\ln x}}{2} - \int {\frac{x}{2}} dx \cr & \int x \ln xdx = \frac{{{x^2}\ln x}}{2} - \frac{1}{2}\int x dx \cr & {\text{integrate using the power rule }}\int {{x^n}} dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C \cr & \int x \ln xdx = \frac{{{x^2}}}{2}\ln x - \frac{1}{2}\left( {\frac{{{x^2}}}{2}} \right) + C \cr & {\text{simplifying}} \cr & \int x \ln xdx = \frac{{{x^2}\ln x}}{2} - \frac{{{x^2}}}{4} + C \cr}