Integral von $$$\frac{1}{x \left(5 - x\right)}$$$

Bestimme $$$\int \frac{1}{x \left(5 - x\right)}\, dx$$$.

Partialbruchzerlegung durchführen (die Schritte sind » zu sehen):

$${\color{red}{\int{\frac{1}{x \left(5 - x\right)} d x}}} = {\color{red}{\int{\left(\frac{1}{5 \left(5 - x\right)} + \frac{1}{5 x}\right)d x}}}$$

Gliedweise integrieren:

$${\color{red}{\int{\left(\frac{1}{5 \left(5 - x\right)} + \frac{1}{5 x}\right)d x}}} = {\color{red}{\left(\int{\frac{1}{5 x} d x} + \int{\frac{1}{5 \left(5 - x\right)} d x}\right)}}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=\frac{1}{5}$$$ und $$$f{\left(x \right)} = \frac{1}{x}$$$ an:

$$\int{\frac{1}{5 \left(5 - x\right)} d x} + {\color{red}{\int{\frac{1}{5 x} d x}}} = \int{\frac{1}{5 \left(5 - x\right)} d x} + {\color{red}{\left(\frac{\int{\frac{1}{x} d x}}{5}\right)}}$$

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

$$\int{\frac{1}{5 \left(5 - x\right)} d x} + \frac{{\color{red}{\int{\frac{1}{x} d x}}}}{5} = \int{\frac{1}{5 \left(5 - x\right)} d x} + \frac{{\color{red}{\ln{\left(\left|{x}\right| \right)}}}}{5}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=\frac{1}{5}$$$ und $$$f{\left(x \right)} = \frac{1}{5 - x}$$$ an:

$$\frac{\ln{\left(\left|{x}\right| \right)}}{5} + {\color{red}{\int{\frac{1}{5 \left(5 - x\right)} d x}}} = \frac{\ln{\left(\left|{x}\right| \right)}}{5} + {\color{red}{\left(\frac{\int{\frac{1}{5 - x} d x}}{5}\right)}}$$

Sei $$$u=5 - x$$$.

Dann $$$du=\left(5 - x\right)^{\prime }dx = - dx$$$ (die Schritte sind » zu sehen), und es gilt $$$dx = - du$$$.

Somit,

$$\frac{\ln{\left(\left|{x}\right| \right)}}{5} + \frac{{\color{red}{\int{\frac{1}{5 - x} d x}}}}{5} = \frac{\ln{\left(\left|{x}\right| \right)}}{5} + \frac{{\color{red}{\int{\left(- \frac{1}{u}\right)d u}}}}{5}$$

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

$$\frac{\ln{\left(\left|{x}\right| \right)}}{5} + \frac{{\color{red}{\int{\left(- \frac{1}{u}\right)d u}}}}{5} = \frac{\ln{\left(\left|{x}\right| \right)}}{5} + \frac{{\color{red}{\left(- \int{\frac{1}{u} d u}\right)}}}{5}$$

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

$$\frac{\ln{\left(\left|{x}\right| \right)}}{5} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{5} = \frac{\ln{\left(\left|{x}\right| \right)}}{5} - \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{5}$$

Zur Erinnerung: $$$u=5 - x$$$:

$$\frac{\ln{\left(\left|{x}\right| \right)}}{5} - \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{5} = \frac{\ln{\left(\left|{x}\right| \right)}}{5} - \frac{\ln{\left(\left|{{\color{red}{\left(5 - x\right)}}}\right| \right)}}{5}$$

Daher,

$$\int{\frac{1}{x \left(5 - x\right)} d x} = \frac{\ln{\left(\left|{x}\right| \right)}}{5} - \frac{\ln{\left(\left|{x - 5}\right| \right)}}{5}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\frac{1}{x \left(5 - x\right)} d x} = \frac{\ln{\left(\left|{x}\right| \right)}}{5} - \frac{\ln{\left(\left|{x - 5}\right| \right)}}{5}+C$$

$$$\int \frac{1}{x \left(5 - x\right)}\, dx = \left(\frac{\ln\left(\left|{x}\right|\right)}{5} - \frac{\ln\left(\left|{x - 5}\right|\right)}{5}\right) + C$$$A