Integral von $$$\frac{1}{\sqrt{x^{2} - 4}}$$$

Bestimme $$$\int \frac{1}{\sqrt{x^{2} - 4}}\, dx$$$.

Sei $$$x=2 \cosh{\left(u \right)}$$$.

Dann $$$dx=\left(2 \cosh{\left(u \right)}\right)^{\prime }du = 2 \sinh{\left(u \right)} du$$$ (die Schritte sind » zu sehen).

Somit folgt, dass $$$u=\operatorname{acosh}{\left(\frac{x}{2} \right)}$$$.

Also,

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

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

$$$\frac{1}{\sqrt{4 \cosh^{2}{\left( u \right)} - 4}}=\frac{1}{2 \sqrt{\cosh^{2}{\left( u \right)} - 1}}=\frac{1}{2 \sqrt{\sinh^{2}{\left( u \right)}}}$$$

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

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

Daher,

$${\color{red}{\int{\frac{1}{\sqrt{x^{2} - 4}} d x}}} = {\color{red}{\int{1 d u}}}$$

Wenden Sie die Konstantenregel $$$\int c\, du = c u$$$ mit $$$c=1$$$ an:

$${\color{red}{\int{1 d u}}} = {\color{red}{u}}$$

Zur Erinnerung: $$$u=\operatorname{acosh}{\left(\frac{x}{2} \right)}$$$:

$${\color{red}{u}} = {\color{red}{\operatorname{acosh}{\left(\frac{x}{2} \right)}}}$$

Daher,

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

Fügen Sie die Integrationskonstante hinzu:

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

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