Funktion $$$\frac{1}{x^{3} \sqrt{x^{2} - 1}}$$$ integraali

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

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

Tällöin $$$dx=\left(\cosh{\left(u \right)}\right)^{\prime }du = \sinh{\left(u \right)} du$$$ (ratkaisuvaiheet ovat nähtävissä »).

Lisäksi seuraa, että $$$u=\operatorname{acosh}{\left(x \right)}$$$.

Integroituva funktio muuttuu

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

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

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

Olettamalla, että $$$\sinh{\left( u \right)} \ge 0$$$, saamme seuraavaa:

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

Integraali voidaan kirjoittaa uudelleen muodossa

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

Kirjoita integroituva hyperbolisen sekantin funktiona:

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

Integraalin $$$\int{\operatorname{sech}^{3}{\left(u \right)} d u}$$$ kohdalla käytä osittaisintegrointia $$$\int \operatorname{m} \operatorname{dv} = \operatorname{m}\operatorname{v} - \int \operatorname{v} \operatorname{dm}$$$.

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

Tällöin $$$\operatorname{dm}=\left(\operatorname{sech}{\left(u \right)}\right)^{\prime }du=- \tanh{\left(u \right)} \operatorname{sech}{\left(u \right)} du$$$ (vaiheet ovat nähtävissä ») ja $$$\operatorname{v}=\int{\operatorname{sech}^{2}{\left(u \right)} d u}=\tanh{\left(u \right)}$$$ (vaiheet ovat nähtävissä »).

Näin ollen,

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

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

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

Laajenna:

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

Summan/erotuksen integraali on integraalien summa/erotus:

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

Näin saamme seuraavan yksinkertaisen lineaarisen yhtälön integraalin suhteen:

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

Ratkaisemalla sen saamme, että

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

Kirjoita hyperbolinen sekantti uudelleen eksponentin $$$\operatorname{sech}\left( u \right)=\frac{2}{e^{\left( u \right)}+e^{-\left( u \right)}}$$$ avulla.:

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

Sovella vakiokertoimen sääntöä $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ käyttäen $$$c=2$$$ ja $$$f{\left(u \right)} = \frac{1}{e^{u} + e^{- u}}$$$:

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

Simplify:

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

Olkoon $$$v=e^{u}$$$.

Tällöin $$$dv=\left(e^{u}\right)^{\prime }du = e^{u} du$$$ (vaiheet ovat nähtävissä ») ja saamme, että $$$e^{u} du = dv$$$.

Integraali voidaan kirjoittaa muotoon

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

Funktion $$$\frac{1}{v^{2} + 1}$$$ integraali on $$$\int{\frac{1}{v^{2} + 1} d v} = \operatorname{atan}{\left(v \right)}$$$:

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

Muista, että $$$v=e^{u}$$$:

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

Muista, että $$$u=\operatorname{acosh}{\left(x \right)}$$$:

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

Näin ollen,

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

Lisää integrointivakio:

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

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