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

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

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

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

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

Der Integrand wird zu

$$$\frac{\sqrt{x^{2} + 4}}{x^{4}} = \frac{\sqrt{4 \sinh^{2}{\left( u \right)} + 4}}{16 \sinh^{4}{\left( u \right)}}$$$

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

$$$\frac{\sqrt{4 \sinh^{2}{\left( u \right)} + 4}}{16 \sinh^{4}{\left( u \right)}}=\frac{\sqrt{\sinh^{2}{\left( u \right)} + 1}}{8 \sinh^{4}{\left( u \right)}}=\frac{\sqrt{\cosh^{2}{\left( u \right)}}}{8 \sinh^{4}{\left( u \right)}}$$$

$$$\frac{\sqrt{\cosh^{2}{\left( u \right)}}}{8 \sinh^{4}{\left( u \right)}} = \frac{\cosh{\left( u \right)}}{8 \sinh^{4}{\left( u \right)}}$$$

Daher,

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

Wende die Konstantenfaktorregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ mit $$$c=\frac{1}{4}$$$ und $$$f{\left(u \right)} = \frac{\cosh^{2}{\left(u \right)}}{\sinh^{4}{\left(u \right)}}$$$ an:

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

Multiplizieren Sie Zähler und Nenner mit $$$\frac{1}{\cosh^{4}{\left( u \right)}}$$$ und wandeln Sie $$$\frac{\cosh^{4}{\left( u \right)}}{\sinh^{4}{\left( u \right)}}$$$ in $$$\frac{1}{\tanh^{4}{\left( u \right)}}$$$ um:

$$\frac{{\color{red}{\int{\frac{\cosh^{2}{\left(u \right)}}{\sinh^{4}{\left(u \right)}} d u}}}}{4} = \frac{{\color{red}{\int{\frac{1}{\cosh^{2}{\left(u \right)} \tanh^{4}{\left(u \right)}} d u}}}}{4}$$

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 $$$\operatorname{sech}^{2}{\left(u \right)} du = dv$$$.

Das Integral wird zu

$$\frac{{\color{red}{\int{\frac{1}{\cosh^{2}{\left(u \right)} \tanh^{4}{\left(u \right)}} d u}}}}{4} = \frac{{\color{red}{\int{\frac{1}{v^{4}} d v}}}}{4}$$

Wenden Sie die Potenzregel $$$\int v^{n}\, dv = \frac{v^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ mit $$$n=-4$$$ an:

$$\frac{{\color{red}{\int{\frac{1}{v^{4}} d v}}}}{4}=\frac{{\color{red}{\int{v^{-4} d v}}}}{4}=\frac{{\color{red}{\frac{v^{-4 + 1}}{-4 + 1}}}}{4}=\frac{{\color{red}{\left(- \frac{v^{-3}}{3}\right)}}}{4}=\frac{{\color{red}{\left(- \frac{1}{3 v^{3}}\right)}}}{4}$$

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

$$- \frac{{\color{red}{v}}^{-3}}{12} = - \frac{{\color{red}{\tanh{\left(u \right)}}}^{-3}}{12}$$

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

$$- \frac{\tanh^{-3}{\left({\color{red}{u}} \right)}}{12} = - \frac{\tanh^{-3}{\left({\color{red}{\operatorname{asinh}{\left(\frac{x}{2} \right)}}} \right)}}{12}$$

Daher,

$$\int{\frac{\sqrt{x^{2} + 4}}{x^{4}} d x} = - \frac{2 \left(\frac{x^{2}}{4} + 1\right)^{\frac{3}{2}}}{3 x^{3}}$$

Vereinfachen:

$$\int{\frac{\sqrt{x^{2} + 4}}{x^{4}} d x} = - \frac{\left(x^{2} + 4\right)^{\frac{3}{2}}}{12 x^{3}}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\frac{\sqrt{x^{2} + 4}}{x^{4}} d x} = - \frac{\left(x^{2} + 4\right)^{\frac{3}{2}}}{12 x^{3}}+C$$

$$$\int \frac{\sqrt{x^{2} + 4}}{x^{4}}\, dx = - \frac{\left(x^{2} + 4\right)^{\frac{3}{2}}}{12 x^{3}} + C$$$A