Integraal van $$$\frac{1}{y^{2}}$$$

Bepaal $$$\int \frac{1}{y^{2}}\, dy$$$.

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

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

Dus,

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

Voeg de integratieconstante toe:

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

$$$\int \frac{1}{y^{2}}\, dy = - \frac{1}{y} + C$$$A