Answer
$$\frac{{2{e^3} + 1}}{9}$$
Work Step by Step
$$\eqalign{
& \int_1^e {{x^2}\ln x} dx \cr
& u = \ln x,{\text{ }}du = \frac{1}{x}dx \cr
& dx = {x^2}dx,{\text{ }}v = \frac{{{x^3}}}{3} \cr
& {\text{integration by parts }}\int {uvdv} = uv - \int {vdu} \cr
& \int {{x^2}\ln x} dx = \frac{{{x^3}}}{3}\ln x - \int {\left( {\frac{{{x^3}}}{3}} \right)\left( {\frac{1}{x}} \right)} dx \cr
& \int {{x^2}\ln x} dx = \frac{{{x^3}}}{3}\ln x - \int {\frac{{{x^2}}}{3}} dx \cr
& \int {{x^2}\ln x} dx = \frac{{{x^3}}}{3}\ln x - \frac{{{x^3}}}{9} + C \cr
& \int_1^e {{x^2}\ln x} dx = \left( {\frac{{{x^3}}}{3}\ln x - \frac{{{x^3}}}{9}} \right)_1^e \cr
& {\text{evaluate limits}} \cr
& = \left( {\frac{{{e^3}}}{3}\ln e - \frac{{{e^3}}}{9}} \right) - \left( {\frac{{{1^3}}}{3}\ln 1 - \frac{{{1^3}}}{9}} \right) \cr
& = \frac{{{e^3}}}{3} - \frac{{{e^3}}}{9} + \frac{1}{9} \cr
& = \frac{{2{e^3} + 1}}{9} \cr} $$
