Integrate x²ln x for x > 0 with u = ln x and dv = x² dx…

Step 1. Parts. $\displaystyle\int x^2\ln x\,dx = \dfrac{x^3}{3}\ln x - \int \dfrac{x^3}{3} \cdot \dfrac{1}{x}\,dx = \dfrac{x^3}{3}\ln x - \dfrac{x^3}{9}$.

Step 2. At $x = 1$. $\ln 1 = 0$, so $F(1) = -\dfrac{1}{9} = -0.1111$.

Step 3. At $x = 2$. $F(2) = \dfrac{8\ln 2}{3} - \dfrac{8}{9} = 1.8484 - 0.8889 = 0.9595$.