Integralen av $$$\sqrt{x^{2} - 2 x + 5}$$$

Bestäm $$$\int \sqrt{x^{2} - 2 x + 5}\, dx$$$.

Kvadratkomplettera (stegen kan ses »): $$$x^{2} - 2 x + 5 = \left(x - 1\right)^{2} + 4$$$:

$${\color{red}{\int{\sqrt{x^{2} - 2 x + 5} d x}}} = {\color{red}{\int{\sqrt{\left(x - 1\right)^{2} + 4} d x}}}$$

Låt $$$u=x - 1$$$ vara.

Då $$$du=\left(x - 1\right)^{\prime }dx = 1 dx$$$ (stegen kan ses »), och vi har att $$$dx = du$$$.

Alltså,

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

Låt $$$u=2 \sinh{\left(v \right)}$$$ vara.

Då $$$du=\left(2 \sinh{\left(v \right)}\right)^{\prime }dv = 2 \cosh{\left(v \right)} dv$$$ (stegen kan ses »).

Det följer också att $$$v=\operatorname{asinh}{\left(\frac{u}{2} \right)}$$$.

Alltså,

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

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

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

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

Alltså,

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

Tillämpa konstantfaktorregeln $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ med $$$c=4$$$ och $$$f{\left(v \right)} = \cosh^{2}{\left(v \right)}$$$:

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

Använd potensreduceringsformeln $$$\cosh^{2}{\left(\alpha \right)} = \frac{\cosh{\left(2 \alpha \right)}}{2} + \frac{1}{2}$$$ med $$$\alpha= v $$$:

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

Tillämpa konstantfaktorregeln $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ med $$$c=\frac{1}{2}$$$ och $$$f{\left(v \right)} = \cosh{\left(2 v \right)} + 1$$$:

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

Integrera termvis:

$$2 {\color{red}{\int{\left(\cosh{\left(2 v \right)} + 1\right)d v}}} = 2 {\color{red}{\left(\int{1 d v} + \int{\cosh{\left(2 v \right)} d v}\right)}}$$

Tillämpa konstantregeln $$$\int c\, dv = c v$$$ med $$$c=1$$$:

$$2 \int{\cosh{\left(2 v \right)} d v} + 2 {\color{red}{\int{1 d v}}} = 2 \int{\cosh{\left(2 v \right)} d v} + 2 {\color{red}{v}}$$

Låt $$$w=2 v$$$ vara.

Då $$$dw=\left(2 v\right)^{\prime }dv = 2 dv$$$ (stegen kan ses »), och vi har att $$$dv = \frac{dw}{2}$$$.

Alltså,

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

Tillämpa konstantfaktorregeln $$$\int c f{\left(w \right)}\, dw = c \int f{\left(w \right)}\, dw$$$ med $$$c=\frac{1}{2}$$$ och $$$f{\left(w \right)} = \cosh{\left(w \right)}$$$:

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

Integralen av den hyperboliska cosinusfunktionen är $$$\int{\cosh{\left(w \right)} d w} = \sinh{\left(w \right)}$$$:

$$2 v + {\color{red}{\int{\cosh{\left(w \right)} d w}}} = 2 v + {\color{red}{\sinh{\left(w \right)}}}$$

Kom ihåg att $$$w=2 v$$$:

$$2 v + \sinh{\left({\color{red}{w}} \right)} = 2 v + \sinh{\left({\color{red}{\left(2 v\right)}} \right)}$$

Kom ihåg att $$$v=\operatorname{asinh}{\left(\frac{u}{2} \right)}$$$:

$$\sinh{\left(2 {\color{red}{v}} \right)} + 2 {\color{red}{v}} = \sinh{\left(2 {\color{red}{\operatorname{asinh}{\left(\frac{u}{2} \right)}}} \right)} + 2 {\color{red}{\operatorname{asinh}{\left(\frac{u}{2} \right)}}}$$

Kom ihåg att $$$u=x - 1$$$:

$$\sinh{\left(2 \operatorname{asinh}{\left(\frac{{\color{red}{u}}}{2} \right)} \right)} + 2 \operatorname{asinh}{\left(\frac{{\color{red}{u}}}{2} \right)} = \sinh{\left(2 \operatorname{asinh}{\left(\frac{{\color{red}{\left(x - 1\right)}}}{2} \right)} \right)} + 2 \operatorname{asinh}{\left(\frac{{\color{red}{\left(x - 1\right)}}}{2} \right)}$$

Alltså,

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

Använd formlerna $$$\sin{\left(2 \operatorname{asin}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{1 - \alpha^{2}}$$$, $$$\sin{\left(2 \operatorname{acos}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{1 - \alpha^{2}}$$$, $$$\cos{\left(2 \operatorname{asin}{\left(\alpha \right)} \right)} = 1 - 2 \alpha^{2}$$$, $$$\cos{\left(2 \operatorname{acos}{\left(\alpha \right)} \right)} = 2 \alpha^{2} - 1$$$, $$$\sinh{\left(2 \operatorname{asinh}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{\alpha^{2} + 1}$$$, $$$\sinh{\left(2 \operatorname{acosh}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{\alpha - 1} \sqrt{\alpha + 1}$$$, $$$\cosh{\left(2 \operatorname{asinh}{\left(\alpha \right)} \right)} = 2 \alpha^{2} + 1$$$, $$$\cosh{\left(2 \operatorname{acosh}{\left(\alpha \right)} \right)} = 2 \alpha^{2} - 1$$$ för att förenkla uttrycket:

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

Förenkla ytterligare:

$$\int{\sqrt{x^{2} - 2 x + 5} d x} = \frac{\left(x - 1\right) \sqrt{\left(x - 1\right)^{2} + 4}}{2} + 2 \operatorname{asinh}{\left(\frac{x}{2} - \frac{1}{2} \right)}$$

Lägg till integrationskonstanten:

$$\int{\sqrt{x^{2} - 2 x + 5} d x} = \frac{\left(x - 1\right) \sqrt{\left(x - 1\right)^{2} + 4}}{2} + 2 \operatorname{asinh}{\left(\frac{x}{2} - \frac{1}{2} \right)}+C$$

$$$\int \sqrt{x^{2} - 2 x + 5}\, dx = \left(\frac{\left(x - 1\right) \sqrt{\left(x - 1\right)^{2} + 4}}{2} + 2 \operatorname{asinh}{\left(\frac{x}{2} - \frac{1}{2} \right)}\right) + C$$$A