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

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

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

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

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

Daher,

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

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

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

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

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

Also,

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

Schreibe in Abhängigkeit vom hyperbolischen Tangens um:

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

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

Dann $$$dv=\left(\tanh{\left(u \right)}\right)^{\prime }du = \operatorname{sech}^{2}{\left(u \right)} du$$$ (die Schritte sind » zu sehen), und es gilt $$$\operatorname{sech}^{2}{\left(u \right)} du = dv$$$.

Also,

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

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

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

Forme den Bruch um und zerlege ihn:

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

Gliedweise integrieren:

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

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

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

Partialbruchzerlegung durchführen (die Schritte sind » zu sehen):

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

Gliedweise integrieren:

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

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

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

Sei $$$w=v - 1$$$.

Dann $$$dw=\left(v - 1\right)^{\prime }dv = 1 dv$$$ (die Schritte sind » zu sehen), und es gilt $$$dv = dw$$$.

Das Integral wird zu

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

Das Integral von $$$\frac{1}{w}$$$ ist $$$\int{\frac{1}{w} d w} = \ln{\left(\left|{w}\right| \right)}$$$:

$$- v + \int{\frac{1}{2 \left(v + 1\right)} d v} - \frac{{\color{red}{\int{\frac{1}{w} d w}}}}{2} = - v + \int{\frac{1}{2 \left(v + 1\right)} d v} - \frac{{\color{red}{\ln{\left(\left|{w}\right| \right)}}}}{2}$$

Zur Erinnerung: $$$w=v - 1$$$:

$$- v - \frac{\ln{\left(\left|{{\color{red}{w}}}\right| \right)}}{2} + \int{\frac{1}{2 \left(v + 1\right)} d v} = - v - \frac{\ln{\left(\left|{{\color{red}{\left(v - 1\right)}}}\right| \right)}}{2} + \int{\frac{1}{2 \left(v + 1\right)} d v}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ mit $$$c=\frac{1}{2}$$$ und $$$f{\left(v \right)} = \frac{1}{v + 1}$$$ an:

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

Sei $$$w=v + 1$$$.

Dann $$$dw=\left(v + 1\right)^{\prime }dv = 1 dv$$$ (die Schritte sind » zu sehen), und es gilt $$$dv = dw$$$.

Das Integral wird zu

$$- v - \frac{\ln{\left(\left|{v - 1}\right| \right)}}{2} + \frac{{\color{red}{\int{\frac{1}{v + 1} d v}}}}{2} = - v - \frac{\ln{\left(\left|{v - 1}\right| \right)}}{2} + \frac{{\color{red}{\int{\frac{1}{w} d w}}}}{2}$$

Das Integral von $$$\frac{1}{w}$$$ ist $$$\int{\frac{1}{w} d w} = \ln{\left(\left|{w}\right| \right)}$$$:

$$- v - \frac{\ln{\left(\left|{v - 1}\right| \right)}}{2} + \frac{{\color{red}{\int{\frac{1}{w} d w}}}}{2} = - v - \frac{\ln{\left(\left|{v - 1}\right| \right)}}{2} + \frac{{\color{red}{\ln{\left(\left|{w}\right| \right)}}}}{2}$$

Zur Erinnerung: $$$w=v + 1$$$:

$$- v - \frac{\ln{\left(\left|{v - 1}\right| \right)}}{2} + \frac{\ln{\left(\left|{{\color{red}{w}}}\right| \right)}}{2} = - v - \frac{\ln{\left(\left|{v - 1}\right| \right)}}{2} + \frac{\ln{\left(\left|{{\color{red}{\left(v + 1\right)}}}\right| \right)}}{2}$$

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

$$- \frac{\ln{\left(\left|{-1 + {\color{red}{v}}}\right| \right)}}{2} + \frac{\ln{\left(\left|{1 + {\color{red}{v}}}\right| \right)}}{2} - {\color{red}{v}} = - \frac{\ln{\left(\left|{-1 + {\color{red}{\tanh{\left(u \right)}}}}\right| \right)}}{2} + \frac{\ln{\left(\left|{1 + {\color{red}{\tanh{\left(u \right)}}}}\right| \right)}}{2} - {\color{red}{\tanh{\left(u \right)}}}$$

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

$$- \frac{\ln{\left(\left|{-1 + \tanh{\left({\color{red}{u}} \right)}}\right| \right)}}{2} + \frac{\ln{\left(\left|{1 + \tanh{\left({\color{red}{u}} \right)}}\right| \right)}}{2} - \tanh{\left({\color{red}{u}} \right)} = - \frac{\ln{\left(\left|{-1 + \tanh{\left({\color{red}{\operatorname{acosh}{\left(x \right)}}} \right)}}\right| \right)}}{2} + \frac{\ln{\left(\left|{1 + \tanh{\left({\color{red}{\operatorname{acosh}{\left(x \right)}}} \right)}}\right| \right)}}{2} - \tanh{\left({\color{red}{\operatorname{acosh}{\left(x \right)}}} \right)}$$

Daher,

$$\int{\frac{\sqrt{x^{2} - 1}}{x^{2}} d x} = - \frac{\ln{\left(\left|{1 - \frac{\sqrt{x - 1} \sqrt{x + 1}}{x}}\right| \right)}}{2} + \frac{\ln{\left(\left|{1 + \frac{\sqrt{x - 1} \sqrt{x + 1}}{x}}\right| \right)}}{2} - \frac{\sqrt{x - 1} \sqrt{x + 1}}{x}$$

Vereinfachen:

$$\int{\frac{\sqrt{x^{2} - 1}}{x^{2}} d x} = \frac{\frac{x \left(- \ln{\left(\left|{\frac{x - \sqrt{x - 1} \sqrt{x + 1}}{x}}\right| \right)} + \ln{\left(\left|{\frac{x + \sqrt{x - 1} \sqrt{x + 1}}{x}}\right| \right)}\right)}{2} - \sqrt{x - 1} \sqrt{x + 1}}{x}$$

Fügen Sie die Integrationskonstante hinzu:

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

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