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

Bestimme $$$\int \frac{1}{u^{2} - 2 u}\, du$$$.

Partialbruchzerlegung durchführen (die Schritte sind » zu sehen):

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

Gliedweise integrieren:

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

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

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

Sei $$$v=u - 2$$$.

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

Das Integral wird zu

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

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

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

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

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

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

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

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

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

Daher,

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

Vereinfachen:

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

Fügen Sie die Integrationskonstante hinzu:

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

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