Integral of $$$t e^{t}$$$

For the integral $$$\int{t e^{t} d t}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Let $$$\operatorname{u}=t$$$ and $$$\operatorname{dv}=e^{t} dt$$$.

Then $$$\operatorname{du}=\left(t\right)^{\prime }dt=1 dt$$$ (steps can be seen ») and $$$\operatorname{v}=\int{e^{t} d t}=e^{t}$$$ (steps can be seen »).

So,

$${\color{red}{\int{t e^{t} d t}}}={\color{red}{\left(t \cdot e^{t}-\int{e^{t} \cdot 1 d t}\right)}}={\color{red}{\left(t e^{t} - \int{e^{t} d t}\right)}}$$

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

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

Therefore,

$$\int{t e^{t} d t} = t e^{t} - e^{t}$$

Simplify:

$$\int{t e^{t} d t} = \left(t - 1\right) e^{t}$$

Add the constant of integration:

$$\int{t e^{t} d t} = \left(t - 1\right) e^{t}+C$$

$$$\int t e^{t}\, dt = \left(t - 1\right) e^{t} + C$$$A