Integral von $$$\frac{1}{\left(y - 1\right)^{2}}$$$

Bestimme $$$\int \frac{1}{\left(y - 1\right)^{2}}\, dy$$$.

Sei $$$u=y - 1$$$.

Dann $$$du=\left(y - 1\right)^{\prime }dy = 1 dy$$$ (die Schritte sind » zu sehen), und es gilt $$$dy = du$$$.

Daher,

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

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

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

Zur Erinnerung: $$$u=y - 1$$$:

$$- {\color{red}{u}}^{-1} = - {\color{red}{\left(y - 1\right)}}^{-1}$$

Daher,

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

Fügen Sie die Integrationskonstante hinzu:

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

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