Integraal van $$$\frac{1}{16 y^{2}}$$$

Bepaal $$$\int \frac{1}{16 y^{2}}\, dy$$$.

Pas de constante-veelvoudregel $$$\int c f{\left(y \right)}\, dy = c \int f{\left(y \right)}\, dy$$$ toe met $$$c=\frac{1}{16}$$$ en $$$f{\left(y \right)} = \frac{1}{y^{2}}$$$:

$${\color{red}{\int{\frac{1}{16 y^{2}} d y}}} = {\color{red}{\left(\frac{\int{\frac{1}{y^{2}} d y}}{16}\right)}}$$

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

$$\frac{{\color{red}{\int{\frac{1}{y^{2}} d y}}}}{16}=\frac{{\color{red}{\int{y^{-2} d y}}}}{16}=\frac{{\color{red}{\frac{y^{-2 + 1}}{-2 + 1}}}}{16}=\frac{{\color{red}{\left(- y^{-1}\right)}}}{16}=\frac{{\color{red}{\left(- \frac{1}{y}\right)}}}{16}$$

Dus,

$$\int{\frac{1}{16 y^{2}} d y} = - \frac{1}{16 y}$$

Voeg de integratieconstante toe:

$$\int{\frac{1}{16 y^{2}} d y} = - \frac{1}{16 y}+C$$

$$$\int \frac{1}{16 y^{2}}\, dy = - \frac{1}{16 y} + C$$$A