Integrate - 1

$\boxed{1.-}$ Relevant wiki: Integration of logarithmic functions

Integrating by parts : $\displaystyle \large \int_{1}^e x^2 \ln x dx = \frac{1}{3}x^3 \ln x\big|_{1}^e - \int_{1}^e \frac{1}{3}x^2 =$ $= \large \frac{e^3}{3} - \frac{1}{9} x^3 \big|_{1}^e = \frac{e^3}{3} - \frac{1}{9} (e^3 - 1) = \frac{1 + 2e^3}{9}...$

$\boxed{2.-}$ Relevant wiki: Integration tricks

Let's call $(1)$ to $I(a)$ $\small (a \neq -1)$ $I(a) = \int_{1}^e x^a dx = \frac{x^{a + 1}}{a + 1}\big|_{1}^e = \frac{e^{a +1}}{a +1} - \frac{1}{a + 1},\space \Rightarrow$ $\frac{\partial }{\partial a} I(a) = \int_{1}^e \frac{\partial }{\partial a} x^a dx = \int_{1}^e x^a \ln x dx =$ derivating $(1)$ $= \frac{e^{a +1}(a + 1) - e^{a +1}}{(a + 1)^2} + \frac{1}{(a + 1)^2} = \frac{ae^{a +1} + 1}{(a + 1)^2} \Rightarrow$ $\frac{\partial}{\partial a} I(2) = \int_{1}^e x^2 \ln x dx = \frac{2e^3 + 1}{9} ...$

Let $u=\ln{x} \implies x \cdot du=dx$ :

$\int_{1}^{e} x^2 \ln{x} \: dx=\int_{0}^{1} x^2 \ln{x} \cdot x \: du=\int_{0}^{1} e^{3u} u \: du$

$\cdots=\left [\dfrac{e^{3u} u}{3} \right ]_{0}^{1}-\int_{0}^{1} \dfrac{e^{3u}}{3} \: du=\dfrac{e^3}{3}-\left [\dfrac{e^{3u}}{9} \right]_{0}{1}=\dfrac{e^3}{3}-\dfrac{e^3-1}{9}$

$\cdots=\dfrac{1+2e^3}{9} \implies A=1,B=9,C=2,D=3 \implies A+B+C+D=15$

$\color{#20A900}{\boxed{\boxed{15}}}$