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

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

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

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

$${\color{red}{\int{\frac{\sqrt{2}}{\sqrt{x}} d x}}} = {\color{red}{\sqrt{2} \int{\frac{1}{\sqrt{x}} d x}}}$$

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

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

Dus,

$$\int{\frac{\sqrt{2}}{\sqrt{x}} d x} = 2 \sqrt{2} \sqrt{x}$$

Voeg de integratieconstante toe:

$$\int{\frac{\sqrt{2}}{\sqrt{x}} d x} = 2 \sqrt{2} \sqrt{x}+C$$

$$$\int \sqrt{2} \sqrt{\frac{1}{x}}\, dx = 2 \sqrt{2} \sqrt{x} + C$$$A