Integral von $$$\frac{\sqrt{4 x^{2} - 25}}{x}$$$

Bestimme $$$\int \frac{\sqrt{4 x^{2} - 25}}{x}\, dx$$$.

Sei $$$x=\frac{5 \cosh{\left(u \right)}}{2}$$$.

Dann $$$dx=\left(\frac{5 \cosh{\left(u \right)}}{2}\right)^{\prime }du = \frac{5 \sinh{\left(u \right)}}{2} du$$$ (die Schritte sind » zu sehen).

Somit folgt, dass $$$u=\operatorname{acosh}{\left(\frac{2 x}{5} \right)}$$$.

Daher,

$$$\frac{\sqrt{4 x^{2} - 25}}{x} = \frac{2 \sqrt{25 \cosh^{2}{\left( u \right)} - 25}}{5 \cosh{\left( u \right)}}$$$

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

$$$\frac{2 \sqrt{25 \cosh^{2}{\left( u \right)} - 25}}{5 \cosh{\left( u \right)}}=\frac{2 \sqrt{\cosh^{2}{\left( u \right)} - 1}}{\cosh{\left( u \right)}}=\frac{2 \sqrt{\sinh^{2}{\left( u \right)}}}{\cosh{\left( u \right)}}$$$

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

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

Also,

$${\color{red}{\int{\frac{\sqrt{4 x^{2} - 25}}{x} d x}}} = {\color{red}{\int{\frac{5 \sinh^{2}{\left(u \right)}}{\cosh{\left(u \right)}} d u}}}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=5$$$ und $$$f{\left(u \right)} = \frac{\sinh^{2}{\left(u \right)}}{\cosh{\left(u \right)}}$$$ an:

$${\color{red}{\int{\frac{5 \sinh^{2}{\left(u \right)}}{\cosh{\left(u \right)}} d u}}} = {\color{red}{\left(5 \int{\frac{\sinh^{2}{\left(u \right)}}{\cosh{\left(u \right)}} d u}\right)}}$$

Multiplizieren Sie Zähler und Nenner mit einem hyperbolischen Kosinus und drücken Sie alles andere in Abhängigkeit vom hyperbolischen Sinus aus, mithilfe der Formel $$$\cosh^2\left(\alpha \right)=\sinh^2\left(\alpha \right)+1$$$ mit $$$\alpha= u $$$:

$$5 {\color{red}{\int{\frac{\sinh^{2}{\left(u \right)}}{\cosh{\left(u \right)}} d u}}} = 5 {\color{red}{\int{\frac{\sinh^{2}{\left(u \right)} \cosh{\left(u \right)}}{\sinh^{2}{\left(u \right)} + 1} d u}}}$$

Sei $$$v=\sinh{\left(u \right)}$$$.

Dann $$$dv=\left(\sinh{\left(u \right)}\right)^{\prime }du = \cosh{\left(u \right)} du$$$ (die Schritte sind » zu sehen), und es gilt $$$\cosh{\left(u \right)} du = dv$$$.

Daher,

$$5 {\color{red}{\int{\frac{\sinh^{2}{\left(u \right)} \cosh{\left(u \right)}}{\sinh^{2}{\left(u \right)} + 1} d u}}} = 5 {\color{red}{\int{\frac{v^{2}}{v^{2} + 1} d v}}}$$

Forme den Bruch um und zerlege ihn:

$$5 {\color{red}{\int{\frac{v^{2}}{v^{2} + 1} d v}}} = 5 {\color{red}{\int{\left(1 - \frac{1}{v^{2} + 1}\right)d v}}}$$

Gliedweise integrieren:

$$5 {\color{red}{\int{\left(1 - \frac{1}{v^{2} + 1}\right)d v}}} = 5 {\color{red}{\left(\int{1 d v} - \int{\frac{1}{v^{2} + 1} d v}\right)}}$$

Wenden Sie die Konstantenregel $$$\int c\, dv = c v$$$ mit $$$c=1$$$ an:

$$- 5 \int{\frac{1}{v^{2} + 1} d v} + 5 {\color{red}{\int{1 d v}}} = - 5 \int{\frac{1}{v^{2} + 1} d v} + 5 {\color{red}{v}}$$

Das Integral von $$$\frac{1}{v^{2} + 1}$$$ ist $$$\int{\frac{1}{v^{2} + 1} d v} = \operatorname{atan}{\left(v \right)}$$$:

$$5 v - 5 {\color{red}{\int{\frac{1}{v^{2} + 1} d v}}} = 5 v - 5 {\color{red}{\operatorname{atan}{\left(v \right)}}}$$

Zur Erinnerung: $$$v=\sinh{\left(u \right)}$$$:

$$- 5 \operatorname{atan}{\left({\color{red}{v}} \right)} + 5 {\color{red}{v}} = - 5 \operatorname{atan}{\left({\color{red}{\sinh{\left(u \right)}}} \right)} + 5 {\color{red}{\sinh{\left(u \right)}}}$$

Zur Erinnerung: $$$u=\operatorname{acosh}{\left(\frac{2 x}{5} \right)}$$$:

$$5 \sinh{\left({\color{red}{u}} \right)} - 5 \operatorname{atan}{\left(\sinh{\left({\color{red}{u}} \right)} \right)} = 5 \sinh{\left({\color{red}{\operatorname{acosh}{\left(\frac{2 x}{5} \right)}}} \right)} - 5 \operatorname{atan}{\left(\sinh{\left({\color{red}{\operatorname{acosh}{\left(\frac{2 x}{5} \right)}}} \right)} \right)}$$

Daher,

$$\int{\frac{\sqrt{4 x^{2} - 25}}{x} d x} = 5 \sqrt{\frac{2 x}{5} - 1} \sqrt{\frac{2 x}{5} + 1} - 5 \operatorname{atan}{\left(\sqrt{\frac{2 x}{5} - 1} \sqrt{\frac{2 x}{5} + 1} \right)}$$

Vereinfachen:

$$\int{\frac{\sqrt{4 x^{2} - 25}}{x} d x} = \sqrt{2 x - 5} \sqrt{2 x + 5} - 5 \operatorname{atan}{\left(\frac{\sqrt{2 x - 5} \sqrt{2 x + 5}}{5} \right)}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\frac{\sqrt{4 x^{2} - 25}}{x} d x} = \sqrt{2 x - 5} \sqrt{2 x + 5} - 5 \operatorname{atan}{\left(\frac{\sqrt{2 x - 5} \sqrt{2 x + 5}}{5} \right)}+C$$

$$$\int \frac{\sqrt{4 x^{2} - 25}}{x}\, dx = \left(\sqrt{2 x - 5} \sqrt{2 x + 5} - 5 \operatorname{atan}{\left(\frac{\sqrt{2 x - 5} \sqrt{2 x + 5}}{5} \right)}\right) + C$$$A