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

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

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

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

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

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

Daher,

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

Fügen Sie die Integrationskonstante hinzu:

$$\int{\frac{1}{a^{2} u^{2}} d u} = - \frac{1}{a^{2} u}+C$$

$$$\int \frac{1}{a^{2} u^{2}}\, du = - \frac{1}{a^{2} u} + C$$$A