[tex]t=\ln x,\ dt=\dfrac{1}{x}\, dx\\du=x^2\,dx,\ u=\dfrac{x^3}{3} \\\\\displaystyle\\\int x^2\ln x\, dx=\dfrac{x^3}{3}\cdot \ln x-\int \dfrac{x^3}{3}\cdot \dfrac{1}{x}\, dx=\dfrac{1}{3}x^3\ln x-\dfrac{1}{3}\int x^2\, dx=\\=\dfrac{1}{3}x^3\ln x-\dfrac{1}{3}\cdot \dfrac{x^3}{3}+C=\dfrac{1}{3}x^3\ln x-\dfrac{x^3}{9}+C[/tex]
[tex]\displaystyle\\\int \limits_1^ex^2\ln x\, dx=\left(\dfrac{1}{3}\cdot e^3\ln e-\dfrac{e^3}{9}\right)-\left(\dfrac{1}{3}\cdot 1^3\ln 1-\dfrac{1^3}{9}\right)=\\\left(\dfrac{e^3}{3}-\dfrac{e^3}{9}\right)-\left(-\dfrac{1}{9}\right)=\left(\dfrac{3e^3}{9}-\dfrac{e^3}{9}\right)+\dfrac{1}{9}=\dfrac{3e^3-e^3}{9}+\dfrac{1}{9}=\dfrac{2e^3+1}{9}[/tex]
