Integral of $$$1 - e^{x}$$$ with respect to $$$e$$$

Find $$$\int \left(1 - e^{x}\right)\, de$$$.

Integrate term by term:

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

Apply the constant rule $$$\int c\, de = c e$$$ with $$$c=1$$$:

$$- \int{e^{x} d e} + {\color{red}{\int{1 d e}}} = - \int{e^{x} d e} + {\color{red}{e}}$$

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

$$e - {\color{red}{\int{e^{x} d e}}}=e - {\color{red}{\frac{e^{x + 1}}{x + 1}}}=e - {\color{red}{\frac{e^{x + 1}}{x + 1}}}$$

Therefore,

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

Simplify:

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

Add the constant of integration:

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

$$$\int \left(1 - e^{x}\right)\, de = \frac{e \left(x + 1\right) - e^{x + 1}}{x + 1} + C$$$A