Answer
$$\frac{{{x^2}}}{2}\ln x - \frac{{{x^2}}}{4} + C$$
Work Step by Step
$$\eqalign{
& \int {x\ln x} dx \cr
& {\text{substitute }}u = \ln x,{\text{ }}du = \frac{1}{x}dx \cr
& dv = xdx,{\text{ }}v = \frac{{{x^2}}}{2} \cr
& {\text{ integration by parts}} \cr
& \int {udv} = uv - \int {vdu} \cr
& {\text{, we have}} \cr
& \int {x\ln x} dx = \ln x\left( {\frac{{{x^2}}}{2}} \right) - \int {\left( {\frac{{{x^2}}}{2}} \right)\left( {\frac{1}{x}} \right)dx} \cr
& \int {x\ln x} dx = \frac{{{x^2}}}{2}\ln x - \int {\frac{x}{2}dx} \cr
& \int {x\ln x} dx = \frac{{{x^2}}}{2}\ln x - \frac{1}{2}\int {xdx} \cr
& {\text{find antiderivative}} \cr
& \int {x\ln x} dx = \frac{{{x^2}}}{2}\ln x - \frac{1}{2}\left( {\frac{{{x^2}}}{2}} \right) + C \cr
& \int {x\ln x} dx = \frac{{{x^2}}}{2}\ln x - \frac{{{x^2}}}{4} + C \cr} $$
