Integral von $$$\frac{x}{\left(1 - x^{2}\right)^{3}}$$$

Bestimme $$$\int \frac{x}{\left(1 - x^{2}\right)^{3}}\, dx$$$.

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,

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

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

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

Wenden Sie die Potenzregel $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ mit $$$n=-3$$$ an:

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

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

$$\frac{{\color{red}{u}}^{-2}}{4} = \frac{{\color{red}{\left(1 - x^{2}\right)}}^{-2}}{4}$$

Daher,

$$\int{\frac{x}{\left(1 - x^{2}\right)^{3}} d x} = \frac{1}{4 \left(1 - x^{2}\right)^{2}}$$

Vereinfachen:

$$\int{\frac{x}{\left(1 - x^{2}\right)^{3}} d x} = \frac{1}{4 \left(x^{2} - 1\right)^{2}}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\frac{x}{\left(1 - x^{2}\right)^{3}} d x} = \frac{1}{4 \left(x^{2} - 1\right)^{2}}+C$$

$$$\int \frac{x}{\left(1 - x^{2}\right)^{3}}\, dx = \frac{1}{4 \left(x^{2} - 1\right)^{2}} + C$$$A