Integraal van $$$\frac{1}{a^{2} v^{2}}$$$ met betrekking tot $$$v$$$

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

Pas de constante-veelvoudregel $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ toe met $$$c=\frac{1}{a^{2}}$$$ en $$$f{\left(v \right)} = \frac{1}{v^{2}}$$$:

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

Pas de machtsregel $$$\int v^{n}\, dv = \frac{v^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ toe met $$$n=-2$$$:

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

Dus,

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

Voeg de integratieconstante toe:

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

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