Integralen av $$$\frac{1}{x^{3} \sqrt{x^{2} - 1}}$$$

Bestäm $$$\int \frac{1}{x^{3} \sqrt{x^{2} - 1}}\, dx$$$.

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

Då $$$dx=\left(\cosh{\left(u \right)}\right)^{\prime }du = \sinh{\left(u \right)} du$$$ (stegen kan ses »).

Det följer också att $$$u=\operatorname{acosh}{\left(x \right)}$$$.

Integranden blir

$$$\frac{1}{x^{3} \sqrt{x^{2} - 1}} = \frac{1}{\sqrt{\cosh^{2}{\left( u \right)} - 1} \cosh^{3}{\left( u \right)}}$$$

Använd identiteten $$$\cosh^{2}{\left( u \right)} - 1 = \sinh^{2}{\left( u \right)}$$$:

$$$\frac{1}{\sqrt{\cosh^{2}{\left( u \right)} - 1} \cosh^{3}{\left( u \right)}}=\frac{1}{\sqrt{\sinh^{2}{\left( u \right)}} \cosh^{3}{\left( u \right)}}$$$

Om vi antar att $$$\sinh{\left( u \right)} \ge 0$$$, erhåller vi följande:

$$$\frac{1}{\sqrt{\sinh^{2}{\left( u \right)}} \cosh^{3}{\left( u \right)}} = \frac{1}{\sinh{\left( u \right)} \cosh^{3}{\left( u \right)}}$$$

Alltså,

$${\color{red}{\int{\frac{1}{x^{3} \sqrt{x^{2} - 1}} d x}}} = {\color{red}{\int{\frac{1}{\cosh^{3}{\left(u \right)}} d u}}}$$

Skriv om integranden i termer av den hyperboliska sekanten:

$${\color{red}{\int{\frac{1}{\cosh^{3}{\left(u \right)}} d u}}} = {\color{red}{\int{\operatorname{sech}^{3}{\left(u \right)} d u}}}$$

För integralen $$$\int{\operatorname{sech}^{3}{\left(u \right)} d u}$$$, använd partiell integration $$$\int \operatorname{m} \operatorname{dv} = \operatorname{m}\operatorname{v} - \int \operatorname{v} \operatorname{dm}$$$.

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

Då gäller $$$\operatorname{dm}=\left(\operatorname{sech}{\left(u \right)}\right)^{\prime }du=- \tanh{\left(u \right)} \operatorname{sech}{\left(u \right)} du$$$ (stegen kan ses ») och $$$\operatorname{v}=\int{\operatorname{sech}^{2}{\left(u \right)} d u}=\tanh{\left(u \right)}$$$ (stegen kan ses »).

Integralen kan omskrivas som

$$\int{\operatorname{sech}^{3}{\left(u \right)} d u}=\operatorname{sech}{\left(u \right)} \cdot \tanh{\left(u \right)}-\int{\tanh{\left(u \right)} \cdot \left(- \tanh{\left(u \right)} \operatorname{sech}{\left(u \right)}\right) d u}=\tanh{\left(u \right)} \operatorname{sech}{\left(u \right)} - \int{\left(- \tanh^{2}{\left(u \right)} \operatorname{sech}{\left(u \right)}\right)d u}$$

Använd formeln $$$\tanh^{2}{\left(u \right)} = 1 - \operatorname{sech}^{2}{\left(u \right)}$$$:

$$\tanh{\left(u \right)} \operatorname{sech}{\left(u \right)} - \int{\left(- \tanh^{2}{\left(u \right)} \operatorname{sech}{\left(u \right)}\right)d u}=\tanh{\left(u \right)} \operatorname{sech}{\left(u \right)} - \int{\left(\operatorname{sech}^{2}{\left(u \right)} - 1\right) \operatorname{sech}{\left(u \right)} d u}$$

Expandera:

$$\tanh{\left(u \right)} \operatorname{sech}{\left(u \right)} - \int{\left(\operatorname{sech}^{2}{\left(u \right)} - 1\right) \operatorname{sech}{\left(u \right)} d u}=\tanh{\left(u \right)} \operatorname{sech}{\left(u \right)} - \int{\left(\operatorname{sech}^{3}{\left(u \right)} - \operatorname{sech}{\left(u \right)}\right)d u}$$

Integralen av en summa/differens är summan/differensen av integraler:

$$\tanh{\left(u \right)} \operatorname{sech}{\left(u \right)} - \int{\left(\operatorname{sech}^{3}{\left(u \right)} - \operatorname{sech}{\left(u \right)}\right)d u}=\tanh{\left(u \right)} \operatorname{sech}{\left(u \right)} + \int{\operatorname{sech}{\left(u \right)} d u} - \int{\operatorname{sech}^{3}{\left(u \right)} d u}$$

Alltså får vi följande enkla linjära ekvation med avseende på integralen:

$${\color{red}{\int{\operatorname{sech}^{3}{\left(u \right)} d u}}}=\tanh{\left(u \right)} \operatorname{sech}{\left(u \right)} + \int{\operatorname{sech}{\left(u \right)} d u} - {\color{red}{\int{\operatorname{sech}^{3}{\left(u \right)} d u}}}$$

Genom att lösa den erhåller vi att

$$\int{\operatorname{sech}^{3}{\left(u \right)} d u}=\frac{\tanh{\left(u \right)} \operatorname{sech}{\left(u \right)}}{2} + \frac{\int{\operatorname{sech}{\left(u \right)} d u}}{2}$$

Skriv om den hyperboliska sekanten med exponenten $$$\operatorname{sech}\left( u \right)=\frac{2}{e^{\left( u \right)}+e^{-\left( u \right)}}$$$:

$$\frac{\tanh{\left(u \right)} \operatorname{sech}{\left(u \right)}}{2} + \frac{{\color{red}{\int{\operatorname{sech}{\left(u \right)} d u}}}}{2} = \frac{\tanh{\left(u \right)} \operatorname{sech}{\left(u \right)}}{2} + \frac{{\color{red}{\int{\frac{2}{e^{u} + e^{- u}} d u}}}}{2}$$

Tillämpa konstantfaktorregeln $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ med $$$c=2$$$ och $$$f{\left(u \right)} = \frac{1}{e^{u} + e^{- u}}$$$:

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

Simplify:

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

Låt $$$v=e^{u}$$$ vara.

Då $$$dv=\left(e^{u}\right)^{\prime }du = e^{u} du$$$ (stegen kan ses »), och vi har att $$$e^{u} du = dv$$$.

Integralen kan omskrivas som

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

Integralen av $$$\frac{1}{v^{2} + 1}$$$ är $$$\int{\frac{1}{v^{2} + 1} d v} = \operatorname{atan}{\left(v \right)}$$$:

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

Kom ihåg att $$$v=e^{u}$$$:

$$\frac{\tanh{\left(u \right)} \operatorname{sech}{\left(u \right)}}{2} + \operatorname{atan}{\left({\color{red}{v}} \right)} = \frac{\tanh{\left(u \right)} \operatorname{sech}{\left(u \right)}}{2} + \operatorname{atan}{\left({\color{red}{e^{u}}} \right)}$$

Kom ihåg att $$$u=\operatorname{acosh}{\left(x \right)}$$$:

$$\operatorname{atan}{\left(e^{{\color{red}{u}}} \right)} + \frac{\operatorname{sech}{\left({\color{red}{u}} \right)} \tanh{\left({\color{red}{u}} \right)}}{2} = \operatorname{atan}{\left(e^{{\color{red}{\operatorname{acosh}{\left(x \right)}}}} \right)} + \frac{\operatorname{sech}{\left({\color{red}{\operatorname{acosh}{\left(x \right)}}} \right)} \tanh{\left({\color{red}{\operatorname{acosh}{\left(x \right)}}} \right)}}{2}$$

Alltså,

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

Lägg till integrationskonstanten:

$$\int{\frac{1}{x^{3} \sqrt{x^{2} - 1}} d x} = \operatorname{atan}{\left(e^{\operatorname{acosh}{\left(x \right)}} \right)} + \frac{\sqrt{x - 1} \sqrt{x + 1}}{2 x^{2}}+C$$

$$$\int \frac{1}{x^{3} \sqrt{x^{2} - 1}}\, dx = \left(\operatorname{atan}{\left(e^{\operatorname{acosh}{\left(x \right)}} \right)} + \frac{\sqrt{x - 1} \sqrt{x + 1}}{2 x^{2}}\right) + C$$$A