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

Find $$$\int t^{2} \ln\left(t\right)\, dt$$$.

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

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

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

So,

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

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

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

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

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

Therefore,

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

Simplify:

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

Add the constant of integration:

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

$$$\int t^{2} \ln\left(t\right)\, dt = \frac{t^{3} \left(3 \ln\left(t\right) - 1\right)}{9} + C$$$A