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

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

Expand the expression:

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

Gliedweise integrieren:

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

Das Integral von $$$\frac{1}{x}$$$ ist $$$\int{\frac{1}{x} d x} = \ln{\left(\left|{x}\right| \right)}$$$:

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

Dieses Integral (Exponentialintegral) besitzt keine geschlossene Form:

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

Daher,

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

Fügen Sie die Integrationskonstante hinzu:

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

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