Integraal van $$$\sqrt{\frac{1}{a x}}$$$ met betrekking tot $$$x$$$

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

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

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

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

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

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

Dus,

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

Voeg de integratieconstante toe:

$$\int{\frac{1}{\sqrt{a} \sqrt{x}} d x} = \frac{2 \sqrt{x}}{\sqrt{a}}+C$$

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