Integral of $$$t e^{2} - 3 e^{t}$$$

Find $$$\int \left(t e^{2} - 3 e^{t}\right)\, dt$$$.

Integrate term by term:

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

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

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

The integral of the exponential function is $$$\int{e^{t} d t} = e^{t}$$$:

$$\int{t e^{2} d t} - 3 {\color{red}{\int{e^{t} d t}}} = \int{t e^{2} d t} - 3 {\color{red}{e^{t}}}$$

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

$$- 3 e^{t} + {\color{red}{\int{t e^{2} d t}}} = - 3 e^{t} + {\color{red}{e^{2} \int{t d t}}}$$

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

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

Therefore,

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

Add the constant of integration:

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

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