Integral of $$$\frac{t - u}{e}$$$ with respect to $$$t$$$

Find $$$\int \frac{t - u}{e}\, dt$$$.

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

$${\color{red}{\int{\frac{t - u}{e} d t}}} = {\color{red}{\frac{\int{\left(t - u\right)d t}}{e}}}$$

Integrate term by term:

$$\frac{{\color{red}{\int{\left(t - u\right)d t}}}}{e} = \frac{{\color{red}{\left(\int{t d t} - \int{u d t}\right)}}}{e}$$

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

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

Apply the constant rule $$$\int c\, dt = c t$$$ with $$$c=u$$$:

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

Therefore,

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

Simplify:

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

Add the constant of integration:

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

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