Integraal van $$$-1 + \frac{1}{v^{2}}$$$

Bepaal $$$\int \left(-1 + \frac{1}{v^{2}}\right)\, dv$$$.

Integreer termgewijs:

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

Pas de constantenregel $$$\int c\, dv = c v$$$ toe met $$$c=1$$$:

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

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

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

Dus,

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

Voeg de integratieconstante toe:

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

$$$\int \left(-1 + \frac{1}{v^{2}}\right)\, dv = \left(- v - \frac{1}{v}\right) + C$$$A