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

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

Let $$$u=e^{x}$$$.

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

So,

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

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 $$$u=e^{x}$$$:

$$e^{{\color{red}{u}}} = e^{{\color{red}{e^{x}}}}$$

Therefore,

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

Add the constant of integration:

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

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