Answer
$$\frac{{2{e^3} + 1}}{9}$$
Work Step by Step
$$\eqalign{
& \int_1^e {{x^2}\ln x} dx \cr
& {\text{substitute }}u = \ln x,{\text{ }}du = \frac{1}{x}dx \cr
& dv = {x^2}dx,{\text{ }}v = \frac{{{x^3}}}{3} \cr
& {\text{applying integration by parts}}{\text{, we have}} \cr
& \int_1^e {{x^2}\ln x} dx = \left. {\left( {\frac{{{x^3}}}{3}\ln x} \right)} \right|_1^e - \int_1^e {\left( {\frac{{{x^3}}}{3}} \right)\left( {\frac{1}{x}dx} \right)} \cr
& \int_1^e {{x^2}\ln x} dx = \left. {\left( {\frac{{{x^3}}}{3}\ln x} \right)} \right|_1^e - \frac{1}{3}\int_1^e {{x^2}dx} \cr
& {\text{integrating}} \cr
& \int_1^e {{x^2}\ln x} dx = \left. {\left( {\frac{{{x^3}}}{3}\ln x} \right)} \right|_1^e - \frac{1}{3}\left. {\left( {\frac{{{x^3}}}{3}} \right)} \right|_1^e \cr
& \int_1^e {{x^2}\ln x} dx = \left. {\left( {\frac{{{x^3}}}{3}\ln x - \frac{{{x^3}}}{9}} \right)} \right|_1^e \cr
& {\text{evaluate limits}} \cr
& = \left( {\frac{{{{\left( e \right)}^3}}}{3}\ln e - \frac{{{{\left( e \right)}^3}}}{9}} \right) - \left( {\frac{{{{\left( 1 \right)}^3}}}{3}\ln 1 - \frac{{{{\left( 1 \right)}^3}}}{9}} \right) \cr
& {\text{simplify}} \cr
& = \frac{{{e^3}}}{3} - \frac{{{e^3}}}{9} + \frac{1}{9} \cr
& = \frac{{2{e^3}}}{9} + \frac{1}{9} \cr
& = \frac{{2{e^3} + 1}}{9} \cr} $$
