Integral of $$$\left(e^{x} + 2\right) e^{- x}$$$

Find $$$\int \left(e^{x} + 2\right) e^{- x}\, dx$$$.

Expand the expression:

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

Integrate term by term:

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

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

$$\int{2 e^{- x} d x} + {\color{red}{\int{1 d x}}} = \int{2 e^{- x} d x} + {\color{red}{x}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=2$$$ and $$$f{\left(x \right)} = e^{- x}$$$:

$$x + {\color{red}{\int{2 e^{- x} d x}}} = x + {\color{red}{\left(2 \int{e^{- x} d x}\right)}}$$

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

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

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

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

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

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

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

$$x - 2 e^{{\color{red}{u}}} = x - 2 e^{{\color{red}{\left(- x\right)}}}$$

Therefore,

$$\int{\left(e^{x} + 2\right) e^{- x} d x} = x - 2 e^{- x}$$

Add the constant of integration:

$$\int{\left(e^{x} + 2\right) e^{- x} d x} = x - 2 e^{- x}+C$$

$$$\int \left(e^{x} + 2\right) e^{- x}\, dx = \left(x - 2 e^{- x}\right) + C$$$A