Integral von $$$\frac{3}{\left(2 - x\right)^{2}}$$$

Bestimme $$$\int \frac{3}{\left(2 - x\right)^{2}}\, dx$$$.

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=3$$$ und $$$f{\left(x \right)} = \frac{1}{\left(2 - x\right)^{2}}$$$ an:

$${\color{red}{\int{\frac{3}{\left(2 - x\right)^{2}} d x}}} = {\color{red}{\left(3 \int{\frac{1}{\left(2 - x\right)^{2}} d x}\right)}}$$

Sei $$$u=2 - x$$$.

Dann $$$du=\left(2 - x\right)^{\prime }dx = - dx$$$ (die Schritte sind » zu sehen), und es gilt $$$dx = - du$$$.

Daher,

$$3 {\color{red}{\int{\frac{1}{\left(2 - x\right)^{2}} d x}}} = 3 {\color{red}{\int{\left(- \frac{1}{u^{2}}\right)d u}}}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=-1$$$ und $$$f{\left(u \right)} = \frac{1}{u^{2}}$$$ an:

$$3 {\color{red}{\int{\left(- \frac{1}{u^{2}}\right)d u}}} = 3 {\color{red}{\left(- \int{\frac{1}{u^{2}} d u}\right)}}$$

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

$$- 3 {\color{red}{\int{\frac{1}{u^{2}} d u}}}=- 3 {\color{red}{\int{u^{-2} d u}}}=- 3 {\color{red}{\frac{u^{-2 + 1}}{-2 + 1}}}=- 3 {\color{red}{\left(- u^{-1}\right)}}=- 3 {\color{red}{\left(- \frac{1}{u}\right)}}$$

Zur Erinnerung: $$$u=2 - x$$$:

$$3 {\color{red}{u}}^{-1} = 3 {\color{red}{\left(2 - x\right)}}^{-1}$$

Daher,

$$\int{\frac{3}{\left(2 - x\right)^{2}} d x} = \frac{3}{2 - x}$$

Vereinfachen:

$$\int{\frac{3}{\left(2 - x\right)^{2}} d x} = - \frac{3}{x - 2}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\frac{3}{\left(2 - x\right)^{2}} d x} = - \frac{3}{x - 2}+C$$

$$$\int \frac{3}{\left(2 - x\right)^{2}}\, dx = - \frac{3}{x - 2} + C$$$A