Integraal van $$$\frac{1}{\sqrt{x^{2} + 4}}$$$

Bepaal $$$\int \frac{1}{\sqrt{x^{2} + 4}}\, dx$$$.

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

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

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

Dus,

$$$\frac{1}{\sqrt{x^{2} + 4}} = \frac{1}{\sqrt{4 \sinh^{2}{\left( u \right)} + 4}}$$$

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

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

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

Dus,

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

Pas de constantenregel $$$\int c\, du = c u$$$ toe met $$$c=1$$$:

$${\color{red}{\int{1 d u}}} = {\color{red}{u}}$$

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

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

Dus,

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

Voeg de integratieconstante toe:

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

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