Integral von $$$\frac{x}{\sqrt{- 6 x^{4} + 2 x^{2}}}$$$

Bestimme $$$\int \frac{x}{\sqrt{- 6 x^{4} + 2 x^{2}}}\, dx$$$.

Die Eingabe wird umgeschrieben: $$$\int{\frac{x}{\sqrt{- 6 x^{4} + 2 x^{2}}} d x}=\int{\frac{1}{\sqrt{2 - 6 x^{2}}} d x}$$$.

Den Integranden vereinfachen:

$${\color{red}{\int{\frac{1}{\sqrt{2 - 6 x^{2}}} d x}}} = {\color{red}{\int{\frac{\sqrt{2}}{2 \sqrt{1 - 3 x^{2}}} d x}}}$$

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

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

Sei $$$x=\frac{\sqrt{3} \sin{\left(u \right)}}{3}$$$.

Dann $$$dx=\left(\frac{\sqrt{3} \sin{\left(u \right)}}{3}\right)^{\prime }du = \frac{\sqrt{3} \cos{\left(u \right)}}{3} du$$$ (die Schritte sind » zu sehen).

Somit folgt, dass $$$u=\operatorname{asin}{\left(\sqrt{3} x \right)}$$$.

Der Integrand wird zu

$$$\frac{1}{\sqrt{1 - 3 x^{2}}} = \frac{1}{\sqrt{1 - \sin^{2}{\left( u \right)}}}$$$

Verwenden Sie die Identität $$$1 - \sin^{2}{\left( u \right)} = \cos^{2}{\left( u \right)}$$$:

$$$\frac{1}{\sqrt{1 - \sin^{2}{\left( u \right)}}}=\frac{1}{\sqrt{\cos^{2}{\left( u \right)}}}$$$

Setzen wir $$$\cos{\left( u \right)} \ge 0$$$ voraus, so erhalten wir Folgendes:

$$$\frac{1}{\sqrt{\cos^{2}{\left( u \right)}}} = \frac{1}{\cos{\left( u \right)}}$$$

Somit,

$$\frac{\sqrt{2} {\color{red}{\int{\frac{1}{\sqrt{1 - 3 x^{2}}} d x}}}}{2} = \frac{\sqrt{2} {\color{red}{\int{\frac{\sqrt{3}}{3} d u}}}}{2}$$

Wenden Sie die Konstantenregel $$$\int c\, du = c u$$$ mit $$$c=\frac{\sqrt{3}}{3}$$$ an:

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

Zur Erinnerung: $$$u=\operatorname{asin}{\left(\sqrt{3} x \right)}$$$:

$$\frac{\sqrt{6} {\color{red}{u}}}{6} = \frac{\sqrt{6} {\color{red}{\operatorname{asin}{\left(\sqrt{3} x \right)}}}}{6}$$

Daher,

$$\int{\frac{1}{\sqrt{2 - 6 x^{2}}} d x} = \frac{\sqrt{6} \operatorname{asin}{\left(\sqrt{3} x \right)}}{6}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\frac{1}{\sqrt{2 - 6 x^{2}}} d x} = \frac{\sqrt{6} \operatorname{asin}{\left(\sqrt{3} x \right)}}{6}+C$$

$$$\int \frac{x}{\sqrt{- 6 x^{4} + 2 x^{2}}}\, dx = \frac{\sqrt{6} \operatorname{asin}{\left(\sqrt{3} x \right)}}{6} + C$$$A