Integral von $$$\frac{1}{\sqrt{a^{2} + x^{2}}}$$$ nach $$$x$$$

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

Sei $$$x=\sinh{\left(u \right)} \left|{a}\right|$$$.

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

Somit folgt, dass $$$u=\operatorname{asinh}{\left(\frac{x}{\left|{a}\right|} \right)}$$$.

Somit,

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

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

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

$$$\frac{1}{\sqrt{\cosh^{2}{\left( u \right)}} \left|{a}\right|} = \frac{1}{\cosh{\left( u \right)} \left|{a}\right|}$$$

Also,

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

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

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

Zur Erinnerung: $$$u=\operatorname{asinh}{\left(\frac{x}{\left|{a}\right|} \right)}$$$:

$${\color{red}{u}} = {\color{red}{\operatorname{asinh}{\left(\frac{x}{\left|{a}\right|} \right)}}}$$

Daher,

$$\int{\frac{1}{\sqrt{a^{2} + x^{2}}} d x} = \operatorname{asinh}{\left(\frac{x}{\left|{a}\right|} \right)}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\frac{1}{\sqrt{a^{2} + x^{2}}} d x} = \operatorname{asinh}{\left(\frac{x}{\left|{a}\right|} \right)}+C$$

$$$\int \frac{1}{\sqrt{a^{2} + x^{2}}}\, dx = \operatorname{asinh}{\left(\frac{x}{\left|{a}\right|} \right)} + C$$$A