Integral von $$$\frac{1}{a^{2} u}$$$ nach $$$u$$$

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

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

$${\color{red}{\int{\frac{1}{a^{2} u} d u}}} = {\color{red}{\frac{\int{\frac{1}{u} d u}}{a^{2}}}}$$

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

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

Daher,

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

Fügen Sie die Integrationskonstante hinzu:

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

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