Integral von $$$\frac{1}{x^{2} \left(x - 1\right)}$$$

Bestimme $$$\int \frac{1}{x^{2} \left(x - 1\right)}\, dx$$$.

Partialbruchzerlegung durchführen (die Schritte sind » zu sehen):

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

Gliedweise integrieren:

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

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

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

Daher,

$$- \int{\frac{1}{x^{2}} d x} - \int{\frac{1}{x} d x} + {\color{red}{\int{\frac{1}{x - 1} d x}}} = - \int{\frac{1}{x^{2}} d x} - \int{\frac{1}{x} d x} + {\color{red}{\int{\frac{1}{u} d u}}}$$

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

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

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

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

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

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

Wenden Sie die Potenzregel $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ mit $$$n=-2$$$ an:

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

Daher,

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

Fügen Sie die Integrationskonstante hinzu:

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

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