Integraal van $$$\frac{\sqrt{\frac{x - 1}{x}}}{x^{2}}$$$

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

De invoer is herschreven: $$$\int{\frac{\sqrt{\frac{x - 1}{x}}}{x^{2}} d x}=\int{\frac{\sqrt{x - 1}}{x^{\frac{5}{2}}} d x}$$$.

Zij $$$u=\sqrt{x}$$$.

Dan $$$du=\left(\sqrt{x}\right)^{\prime }dx = \frac{1}{2 \sqrt{x}} dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$\frac{dx}{\sqrt{x}} = 2 du$$$.

Dus,

$${\color{red}{\int{\frac{\sqrt{x - 1}}{x^{\frac{5}{2}}} d x}}} = {\color{red}{\int{\frac{2 \sqrt{u^{2} - 1}}{u^{4}} d u}}}$$

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

$${\color{red}{\int{\frac{2 \sqrt{u^{2} - 1}}{u^{4}} d u}}} = {\color{red}{\left(2 \int{\frac{\sqrt{u^{2} - 1}}{u^{4}} d u}\right)}}$$

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

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

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

Dus,

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

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

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

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

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

Dus,

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

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

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

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

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

Dus,

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

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

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

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

$$\frac{2 {\color{red}{w}}^{3}}{3} = \frac{2 {\color{red}{\tanh{\left(v \right)}}}^{3}}{3}$$

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

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

We herinneren eraan dat $$$u=\sqrt{x}$$$:

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

Dus,

$$\int{\frac{\sqrt{x - 1}}{x^{\frac{5}{2}}} d x} = \frac{2 \left(\sqrt{x} - 1\right)^{\frac{3}{2}} \left(\sqrt{x} + 1\right)^{\frac{3}{2}}}{3 x^{\frac{3}{2}}}$$

Voeg de integratieconstante toe:

$$\int{\frac{\sqrt{x - 1}}{x^{\frac{5}{2}}} d x} = \frac{2 \left(\sqrt{x} - 1\right)^{\frac{3}{2}} \left(\sqrt{x} + 1\right)^{\frac{3}{2}}}{3 x^{\frac{3}{2}}}+C$$

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