Integral of $$$e^{x + 2}$$$

Find $$$\int e^{x + 2}\, dx$$$.

Let $$$u=x + 2$$$.

Then $$$du=\left(x + 2\right)^{\prime }dx = 1 dx$$$ (steps can be seen »), and we have that $$$dx = du$$$.

The integral can be rewritten as

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

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

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

Recall that $$$u=x + 2$$$:

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

Therefore,

$$\int{e^{x + 2} d x} = e^{x + 2}$$

Add the constant of integration:

$$\int{e^{x + 2} d x} = e^{x + 2}+C$$

$$$\int e^{x + 2}\, dx = e^{x + 2} + C$$$A