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

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

Partialbruchzerlegung durchführen (die Schritte sind » zu sehen):

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

Gliedweise integrieren:

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

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

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

Das Integral wird zu

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

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

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

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

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

Sei $$$u=x - 2$$$.

Dann $$$du=\left(x - 2\right)^{\prime }dx = 1 dx$$$ (die Schritte sind » zu sehen), und es gilt $$$dx = du$$$.

Das Integral wird zu

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

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

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

Zur Erinnerung: $$$u=x - 2$$$:

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

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

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

Sei $$$u=x - 2$$$.

Dann $$$du=\left(x - 2\right)^{\prime }dx = 1 dx$$$ (die Schritte sind » zu sehen), und es gilt $$$dx = du$$$.

Also,

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

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

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

Zur Erinnerung: $$$u=x - 2$$$:

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

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

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

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 wird zu

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

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

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

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

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

Daher,

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

Vereinfachen:

$$\int{\frac{1}{\left(x - 2\right)^{2} \left(x - 1\right)^{2}} d x} = \frac{- 2 x + 2 \left(x - 2\right) \left(x - 1\right) \left(- \ln{\left(\left|{x - 2}\right| \right)} + \ln{\left(\left|{x - 1}\right| \right)}\right) + 3}{\left(x - 2\right) \left(x - 1\right)}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\frac{1}{\left(x - 2\right)^{2} \left(x - 1\right)^{2}} d x} = \frac{- 2 x + 2 \left(x - 2\right) \left(x - 1\right) \left(- \ln{\left(\left|{x - 2}\right| \right)} + \ln{\left(\left|{x - 1}\right| \right)}\right) + 3}{\left(x - 2\right) \left(x - 1\right)}+C$$

$$$\int \frac{1}{\left(x - 2\right)^{2} \left(x - 1\right)^{2}}\, dx = \frac{- 2 x + 2 \left(x - 2\right) \left(x - 1\right) \left(- \ln\left(\left|{x - 2}\right|\right) + \ln\left(\left|{x - 1}\right|\right)\right) + 3}{\left(x - 2\right) \left(x - 1\right)} + C$$$A