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

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

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

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

Dus,

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

Voeg de integratieconstante toe:

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

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