Integral of $$$x^{2} \ln\left(x\right)$$$

For the integral $$$\int{x^{2} \ln{\left(x \right)} d x}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Let $$$\operatorname{u}=\ln{\left(x \right)}$$$ and $$$\operatorname{dv}=x^{2} dx$$$.

Then $$$\operatorname{du}=\left(\ln{\left(x \right)}\right)^{\prime }dx=\frac{dx}{x}$$$ (steps can be seen ») and $$$\operatorname{v}=\int{x^{2} d x}=\frac{x^{3}}{3}$$$ (steps can be seen »).

Thus,

$${\color{red}{\int{x^{2} \ln{\left(x \right)} d x}}}={\color{red}{\left(\ln{\left(x \right)} \cdot \frac{x^{3}}{3}-\int{\frac{x^{3}}{3} \cdot \frac{1}{x} d x}\right)}}={\color{red}{\left(\frac{x^{3} \ln{\left(x \right)}}{3} - \int{\frac{x^{2}}{3} d x}\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{3}$$$ and $$$f{\left(x \right)} = x^{2}$$$:

$$\frac{x^{3} \ln{\left(x \right)}}{3} - {\color{red}{\int{\frac{x^{2}}{3} d x}}} = \frac{x^{3} \ln{\left(x \right)}}{3} - {\color{red}{\left(\frac{\int{x^{2} d x}}{3}\right)}}$$

Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=2$$$:

$$\frac{x^{3} \ln{\left(x \right)}}{3} - \frac{{\color{red}{\int{x^{2} d x}}}}{3}=\frac{x^{3} \ln{\left(x \right)}}{3} - \frac{{\color{red}{\frac{x^{1 + 2}}{1 + 2}}}}{3}=\frac{x^{3} \ln{\left(x \right)}}{3} - \frac{{\color{red}{\left(\frac{x^{3}}{3}\right)}}}{3}$$

Therefore,

$$\int{x^{2} \ln{\left(x \right)} d x} = \frac{x^{3} \ln{\left(x \right)}}{3} - \frac{x^{3}}{9}$$

Simplify:

$$\int{x^{2} \ln{\left(x \right)} d x} = \frac{x^{3} \left(3 \ln{\left(x \right)} - 1\right)}{9}$$

Add the constant of integration:

$$\int{x^{2} \ln{\left(x \right)} d x} = \frac{x^{3} \left(3 \ln{\left(x \right)} - 1\right)}{9}+C$$

Answer: $$$\int{x^{2} \ln{\left(x \right)} d x}=\frac{x^{3} \left(3 \ln{\left(x \right)} - 1\right)}{9}+C$$$