Evaluate the following integral:
$${\int }_1^e{x}^2\ln (x)dx$$

We use Integration by Parts: $\int udv=uv-\int vdu$.Let $u=\ln (x)$ and $dv=x^{2}dx$.Then $du=\frac{1}{x}dx$ and $v=\frac{x^3}{3}$.Applying the formula:
$$I={\left[\ln (x)\cdot \frac{x^3}{3}\right]}_1^e-{\int }_1^e\frac{x^3}{3}\cdot \frac{1}{x}dx$$
$$I={\left[\frac{x^3\ln (x)}{3}\right]}_1^e-\frac{1}{3}{\int }_1^e{x}^{2}dx$$
$$I=\left(\frac{e^3\ln (e)}{3}-\frac{1^3\ln (1)}{3}\right)-\frac{1}{3}{\left[\frac{x^3}{3}\right]}_1^e$$
$$I=\frac{e^3}{3}-0-\frac{1}{9}[e^3-1]$$
$$I=\frac{e^3}{3}-\frac{e^3}{9}+\frac{1}{9}$$
$$I=\frac{3e^3-e^3}{9}+\frac{1}{9}=\frac{2}{9}{e}^3+\frac{1}{9}$$
Note: The options provided in the source material for this specific problem contained slight variations; based on the standard calculation, Option C is the intended answer.