Integral von $$$\frac{1}{\sqrt{1 - x}}$$$

Bestimme $$$\int \frac{1}{\sqrt{1 - x}}\, dx$$$.

Sei $$$u=1 - x$$$.

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

Daher,

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

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

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

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

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

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

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

Daher,

$$\int{\frac{1}{\sqrt{1 - x}} d x} = - 2 \sqrt{1 - x}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\frac{1}{\sqrt{1 - x}} d x} = - 2 \sqrt{1 - x}+C$$

$$$\int \frac{1}{\sqrt{1 - x}}\, dx = - 2 \sqrt{1 - x} + C$$$A