Integral von $$$\frac{1}{p^{2}}$$$

Bestimme $$$\int \frac{1}{p^{2}}\, dp$$$.

Wenden Sie die Potenzregel $$$\int p^{n}\, dp = \frac{p^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ mit $$$n=-2$$$ an:

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

Daher,

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

Fügen Sie die Integrationskonstante hinzu:

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

$$$\int \frac{1}{p^{2}}\, dp = - \frac{1}{p} + C$$$A