Integraal van $$$\frac{1}{4 t^{2}}$$$

Bepaal $$$\int \frac{1}{4 t^{2}}\, dt$$$.

Pas de constante-veelvoudregel $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ toe met $$$c=\frac{1}{4}$$$ en $$$f{\left(t \right)} = \frac{1}{t^{2}}$$$:

$${\color{red}{\int{\frac{1}{4 t^{2}} d t}}} = {\color{red}{\left(\frac{\int{\frac{1}{t^{2}} d t}}{4}\right)}}$$

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

$$\frac{{\color{red}{\int{\frac{1}{t^{2}} d t}}}}{4}=\frac{{\color{red}{\int{t^{-2} d t}}}}{4}=\frac{{\color{red}{\frac{t^{-2 + 1}}{-2 + 1}}}}{4}=\frac{{\color{red}{\left(- t^{-1}\right)}}}{4}=\frac{{\color{red}{\left(- \frac{1}{t}\right)}}}{4}$$

Dus,

$$\int{\frac{1}{4 t^{2}} d t} = - \frac{1}{4 t}$$

Voeg de integratieconstante toe:

$$\int{\frac{1}{4 t^{2}} d t} = - \frac{1}{4 t}+C$$

$$$\int \frac{1}{4 t^{2}}\, dt = - \frac{1}{4 t} + C$$$A