Integral von $$$- \alpha \beta - x + \frac{1}{x}$$$ nach $$$x$$$

Bestimme $$$\int \left(- \alpha \beta - x + \frac{1}{x}\right)\, dx$$$.

Gliedweise integrieren:

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

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

$$- \int{x d x} - \int{\alpha \beta d x} + {\color{red}{\int{\frac{1}{x} d x}}} = - \int{x d x} - \int{\alpha \beta d x} + {\color{red}{\ln{\left(\left|{x}\right| \right)}}}$$

Wenden Sie die Potenzregel $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ mit $$$n=1$$$ an:

$$\ln{\left(\left|{x}\right| \right)} - \int{\alpha \beta d x} - {\color{red}{\int{x d x}}}=\ln{\left(\left|{x}\right| \right)} - \int{\alpha \beta d x} - {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}=\ln{\left(\left|{x}\right| \right)} - \int{\alpha \beta d x} - {\color{red}{\left(\frac{x^{2}}{2}\right)}}$$

Wenden Sie die Konstantenregel $$$\int c\, dx = c x$$$ mit $$$c=\alpha \beta$$$ an:

$$- \frac{x^{2}}{2} + \ln{\left(\left|{x}\right| \right)} - {\color{red}{\int{\alpha \beta d x}}} = - \frac{x^{2}}{2} + \ln{\left(\left|{x}\right| \right)} - {\color{red}{\alpha \beta x}}$$

Daher,

$$\int{\left(- \alpha \beta - x + \frac{1}{x}\right)d x} = - \alpha \beta x - \frac{x^{2}}{2} + \ln{\left(\left|{x}\right| \right)}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\left(- \alpha \beta - x + \frac{1}{x}\right)d x} = - \alpha \beta x - \frac{x^{2}}{2} + \ln{\left(\left|{x}\right| \right)}+C$$

$$$\int \left(- \alpha \beta - x + \frac{1}{x}\right)\, dx = \left(- \alpha \beta x - \frac{x^{2}}{2} + \ln\left(\left|{x}\right|\right)\right) + C$$$A