Integral of $$$-1 + e^{- x}$$$

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

Integrate term by term:

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

Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=1$$$:

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

Let $$$u=- x$$$.

Then $$$du=\left(- x\right)^{\prime }dx = - dx$$$ (steps can be seen »), and we have that $$$dx = - du$$$.

The integral can be rewritten as

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

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

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

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

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

Recall that $$$u=- x$$$:

$$- x - e^{{\color{red}{u}}} = - x - e^{{\color{red}{\left(- x\right)}}}$$

Therefore,

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

Add the constant of integration:

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

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