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

Bestimme $$$\int \frac{1}{a^{2} v^{2}}\, dv$$$.

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

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

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

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

Daher,

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

Fügen Sie die Integrationskonstante hinzu:

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

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