Answer
\[ = \frac{{{x^2}\ln x}}{2} - \frac{{{x^2}}}{4} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {x\ln x\,\,dx} \hfill \\
\hfill \\
set\,\,\,the\,\,substitution \hfill \\
\hfill \\
u = \ln x\,\,\,\,\,\,\, \to \,\,\,\,\,\,du = \frac{{dx}}{x} \hfill \\
dv = xdx\,\,\, \to \,\,\,\,\,\,\,\,v = \frac{{{x^2}}}{2} \hfill \\
\hfill \\
use\,\,uv - \int_{}^{} {vdu} \hfill \\
\hfill \\
{\text{replacing the values }}{\text{in the equation}} \hfill \\
\hfill \\
= \frac{{{x^2}}}{2}\ln x - \int_{}^{} {\,\left( {\frac{{{x^2}}}{2}} \right)\,\left( {\frac{{dx}}{x}} \right)} \hfill \\
\hfill \\
integrate\,\,{\text{using }}\int {{x^n}dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C} \hfill \\
\hfill \\
= \frac{{{x^2}\ln x}}{2} - \frac{{{x^2}}}{4} + C \hfill \\
\end{gathered} \]
