Integral von $$$\frac{1}{x^{2} - 1}$$$

Bestimme $$$\int \frac{1}{x^{2} - 1}\, dx$$$.

Partialbruchzerlegung durchführen (die Schritte sind » zu sehen):

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

Gliedweise integrieren:

$${\color{red}{\int{\left(- \frac{1}{2 \left(x + 1\right)} + \frac{1}{2 \left(x - 1\right)}\right)d x}}} = {\color{red}{\left(\int{\frac{1}{2 \left(x - 1\right)} d x} - \int{\frac{1}{2 \left(x + 1\right)} d x}\right)}}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=\frac{1}{2}$$$ und $$$f{\left(x \right)} = \frac{1}{x - 1}$$$ an:

$$- \int{\frac{1}{2 \left(x + 1\right)} d x} + {\color{red}{\int{\frac{1}{2 \left(x - 1\right)} d x}}} = - \int{\frac{1}{2 \left(x + 1\right)} d x} + {\color{red}{\left(\frac{\int{\frac{1}{x - 1} d x}}{2}\right)}}$$

Sei $$$u=x - 1$$$.

Dann $$$du=\left(x - 1\right)^{\prime }dx = 1 dx$$$ (die Schritte sind » zu sehen), und es gilt $$$dx = du$$$.

Das Integral lässt sich umschreiben als

$$- \int{\frac{1}{2 \left(x + 1\right)} d x} + \frac{{\color{red}{\int{\frac{1}{x - 1} d x}}}}{2} = - \int{\frac{1}{2 \left(x + 1\right)} d x} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2}$$

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

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

Zur Erinnerung: $$$u=x - 1$$$:

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

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=\frac{1}{2}$$$ und $$$f{\left(x \right)} = \frac{1}{x + 1}$$$ an:

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

Sei $$$u=x + 1$$$.

Dann $$$du=\left(x + 1\right)^{\prime }dx = 1 dx$$$ (die Schritte sind » zu sehen), und es gilt $$$dx = du$$$.

Somit,

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

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

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

Zur Erinnerung: $$$u=x + 1$$$:

$$\frac{\ln{\left(\left|{x - 1}\right| \right)}}{2} - \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{2} = \frac{\ln{\left(\left|{x - 1}\right| \right)}}{2} - \frac{\ln{\left(\left|{{\color{red}{\left(x + 1\right)}}}\right| \right)}}{2}$$

Daher,

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

Fügen Sie die Integrationskonstante hinzu:

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

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