Integraal van $$$\frac{\sqrt{x^{2} - 4}}{x^{4}}$$$

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

Zij $$$x=2 \cosh{\left(u \right)}$$$.

Dan $$$dx=\left(2 \cosh{\left(u \right)}\right)^{\prime }du = 2 \sinh{\left(u \right)} du$$$ (zie » voor de stappen).

Bovendien volgt dat $$$u=\operatorname{acosh}{\left(\frac{x}{2} \right)}$$$.

De integraand wordt

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

Gebruik de identiteit $$$\cosh^{2}{\left( u \right)} - 1 = \sinh^{2}{\left( u \right)}$$$:

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

Aangenomen dat $$$\sinh{\left( u \right)} \ge 0$$$, verkrijgen we het volgende:

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

De integraal kan worden herschreven als

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

Pas de constante-veelvoudregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ toe met $$$c=\frac{1}{4}$$$ en $$$f{\left(u \right)} = \frac{\sinh^{2}{\left(u \right)}}{\cosh^{4}{\left(u \right)}}$$$:

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

Vermenigvuldig de teller en de noemer met $$$\cosh^{2}{\left( u \right)}$$$ en zet $$$\frac{\sinh^{2}{\left( u \right)}}{\cosh^{2}{\left( u \right)}}$$$ om in $$$\tanh^{2}{\left( u \right)}$$$:

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

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

Dan $$$dv=\left(\tanh{\left(u \right)}\right)^{\prime }du = \operatorname{sech}^{2}{\left(u \right)} du$$$ (de stappen zijn te zien »), en dan geldt dat $$$\operatorname{sech}^{2}{\left(u \right)} du = dv$$$.

Dus,

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

Pas de machtsregel $$$\int v^{n}\, dv = \frac{v^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ toe met $$$n=2$$$:

$$\frac{{\color{red}{\int{v^{2} d v}}}}{4}=\frac{{\color{red}{\frac{v^{1 + 2}}{1 + 2}}}}{4}=\frac{{\color{red}{\left(\frac{v^{3}}{3}\right)}}}{4}$$

We herinneren eraan dat $$$v=\tanh{\left(u \right)}$$$:

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

We herinneren eraan dat $$$u=\operatorname{acosh}{\left(\frac{x}{2} \right)}$$$:

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

Dus,

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

Vereenvoudig:

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

Voeg de integratieconstante toe:

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

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