Integral von $$$\frac{x}{\sqrt{1 - x^{2}}}$$$

Bestimme $$$\int \frac{x}{\sqrt{1 - x^{2}}}\, dx$$$.

Sei $$$u=1 - x^{2}$$$.

Dann $$$du=\left(1 - x^{2}\right)^{\prime }dx = - 2 x dx$$$ (die Schritte sind » zu sehen), und es gilt $$$x dx = - \frac{du}{2}$$$.

Somit,

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

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

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

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:

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

Zur Erinnerung: $$$u=1 - x^{2}$$$:

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

Daher,

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

Fügen Sie die Integrationskonstante hinzu:

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

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