Integral of $$$\frac{e^{x} - 1}{x}$$$

Find $$$\int \frac{e^{x} - 1}{x}\, dx$$$.

Expand the expression:

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

Integrate term by term:

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

The integral of $$$\frac{1}{x}$$$ is $$$\int{\frac{1}{x} d x} = \ln{\left(\left|{x}\right| \right)}$$$:

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

This integral (Exponential Integral) does not have a closed form:

$$- \ln{\left(\left|{x}\right| \right)} + {\color{red}{\int{\frac{e^{x}}{x} d x}}} = - \ln{\left(\left|{x}\right| \right)} + {\color{red}{\operatorname{Ei}{\left(x \right)}}}$$

Therefore,

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

Add the constant of integration:

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

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