Integral of $$$t e^{\frac{1}{2}}$$$

Find $$$\int t e^{\frac{1}{2}}\, dt$$$.

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

$${\color{red}{\int{t e^{\frac{1}{2}} d t}}} = {\color{red}{e^{\frac{1}{2}} \int{t d t}}}$$

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

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

Therefore,

$$\int{t e^{\frac{1}{2}} d t} = \frac{t^{2} e^{\frac{1}{2}}}{2}$$

Add the constant of integration:

$$\int{t e^{\frac{1}{2}} d t} = \frac{t^{2} e^{\frac{1}{2}}}{2}+C$$

$$$\int t e^{\frac{1}{2}}\, dt = \frac{t^{2} e^{\frac{1}{2}}}{2} + C$$$A