Integraal van $$$\sqrt{x^{2} + 6 x + 10}$$$

Bepaal $$$\int \sqrt{x^{2} + 6 x + 10}\, dx$$$.

Voltooi het kwadraat (stappen zijn te zien »): $$$x^{2} + 6 x + 10 = \left(x + 3\right)^{2} + 1$$$:

$${\color{red}{\int{\sqrt{x^{2} + 6 x + 10} d x}}} = {\color{red}{\int{\sqrt{\left(x + 3\right)^{2} + 1} d x}}}$$

Zij $$$u=x + 3$$$.

Dan $$$du=\left(x + 3\right)^{\prime }dx = 1 dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$dx = du$$$.

Dus,

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

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

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

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

De integraand wordt

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

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

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

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

De integraal kan worden herschreven als

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

Pas de machtsreductieformule $$$\cosh^{2}{\left(\alpha \right)} = \frac{\cosh{\left(2 \alpha \right)}}{2} + \frac{1}{2}$$$ toe met $$$\alpha= v $$$:

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

Pas de constante-veelvoudregel $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ toe met $$$c=\frac{1}{2}$$$ en $$$f{\left(v \right)} = \cosh{\left(2 v \right)} + 1$$$:

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

Integreer termgewijs:

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

Pas de constantenregel $$$\int c\, dv = c v$$$ toe met $$$c=1$$$:

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

Zij $$$w=2 v$$$.

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

Dus,

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

Pas de constante-veelvoudregel $$$\int c f{\left(w \right)}\, dw = c \int f{\left(w \right)}\, dw$$$ toe met $$$c=\frac{1}{2}$$$ en $$$f{\left(w \right)} = \cosh{\left(w \right)}$$$:

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

De integraal van de cosinus hyperbolicus is $$$\int{\cosh{\left(w \right)} d w} = \sinh{\left(w \right)}$$$:

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

We herinneren eraan dat $$$w=2 v$$$:

$$\frac{v}{2} + \frac{\sinh{\left({\color{red}{w}} \right)}}{4} = \frac{v}{2} + \frac{\sinh{\left({\color{red}{\left(2 v\right)}} \right)}}{4}$$

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

$$\frac{\sinh{\left(2 {\color{red}{v}} \right)}}{4} + \frac{{\color{red}{v}}}{2} = \frac{\sinh{\left(2 {\color{red}{\operatorname{asinh}{\left(u \right)}}} \right)}}{4} + \frac{{\color{red}{\operatorname{asinh}{\left(u \right)}}}}{2}$$

We herinneren eraan dat $$$u=x + 3$$$:

$$\frac{\sinh{\left(2 \operatorname{asinh}{\left({\color{red}{u}} \right)} \right)}}{4} + \frac{\operatorname{asinh}{\left({\color{red}{u}} \right)}}{2} = \frac{\sinh{\left(2 \operatorname{asinh}{\left({\color{red}{\left(x + 3\right)}} \right)} \right)}}{4} + \frac{\operatorname{asinh}{\left({\color{red}{\left(x + 3\right)}} \right)}}{2}$$

Dus,

$$\int{\sqrt{x^{2} + 6 x + 10} d x} = \frac{\sinh{\left(2 \operatorname{asinh}{\left(x + 3 \right)} \right)}}{4} + \frac{\operatorname{asinh}{\left(x + 3 \right)}}{2}$$

Gebruik de formules $$$\sin{\left(2 \operatorname{asin}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{1 - \alpha^{2}}$$$, $$$\sin{\left(2 \operatorname{acos}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{1 - \alpha^{2}}$$$, $$$\cos{\left(2 \operatorname{asin}{\left(\alpha \right)} \right)} = 1 - 2 \alpha^{2}$$$, $$$\cos{\left(2 \operatorname{acos}{\left(\alpha \right)} \right)} = 2 \alpha^{2} - 1$$$, $$$\sinh{\left(2 \operatorname{asinh}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{\alpha^{2} + 1}$$$, $$$\sinh{\left(2 \operatorname{acosh}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{\alpha - 1} \sqrt{\alpha + 1}$$$, $$$\cosh{\left(2 \operatorname{asinh}{\left(\alpha \right)} \right)} = 2 \alpha^{2} + 1$$$, $$$\cosh{\left(2 \operatorname{acosh}{\left(\alpha \right)} \right)} = 2 \alpha^{2} - 1$$$ om de uitdrukking te vereenvoudigen:

$$\int{\sqrt{x^{2} + 6 x + 10} d x} = \frac{\left(x + 3\right) \sqrt{\left(x + 3\right)^{2} + 1}}{2} + \frac{\operatorname{asinh}{\left(x + 3 \right)}}{2}$$

Voeg de integratieconstante toe:

$$\int{\sqrt{x^{2} + 6 x + 10} d x} = \frac{\left(x + 3\right) \sqrt{\left(x + 3\right)^{2} + 1}}{2} + \frac{\operatorname{asinh}{\left(x + 3 \right)}}{2}+C$$

$$$\int \sqrt{x^{2} + 6 x + 10}\, dx = \left(\frac{\left(x + 3\right) \sqrt{\left(x + 3\right)^{2} + 1}}{2} + \frac{\operatorname{asinh}{\left(x + 3 \right)}}{2}\right) + C$$$A