Integraal van $$$\sqrt{x} z - \sqrt{3} x$$$ met betrekking tot $$$x$$$

Bepaal $$$\int \left(\sqrt{x} z - \sqrt{3} x\right)\, dx$$$.

Integreer termgewijs:

$${\color{red}{\int{\left(\sqrt{x} z - \sqrt{3} x\right)d x}}} = {\color{red}{\left(- \int{\sqrt{3} x d x} + \int{\sqrt{x} z d x}\right)}}$$

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

$$- \int{\sqrt{3} x d x} + {\color{red}{\int{\sqrt{x} z d x}}} = - \int{\sqrt{3} x d x} + {\color{red}{z \int{\sqrt{x} d x}}}$$

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

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

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=\sqrt{3}$$$ en $$$f{\left(x \right)} = x$$$:

$$\frac{2 x^{\frac{3}{2}} z}{3} - {\color{red}{\int{\sqrt{3} x d x}}} = \frac{2 x^{\frac{3}{2}} z}{3} - {\color{red}{\sqrt{3} \int{x d x}}}$$

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

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

Dus,

$$\int{\left(\sqrt{x} z - \sqrt{3} x\right)d x} = \frac{2 x^{\frac{3}{2}} z}{3} - \frac{\sqrt{3} x^{2}}{2}$$

Voeg de integratieconstante toe:

$$\int{\left(\sqrt{x} z - \sqrt{3} x\right)d x} = \frac{2 x^{\frac{3}{2}} z}{3} - \frac{\sqrt{3} x^{2}}{2}+C$$

$$$\int \left(\sqrt{x} z - \sqrt{3} x\right)\, dx = \left(\frac{2 x^{\frac{3}{2}} z}{3} - \frac{\sqrt{3} x^{2}}{2}\right) + C$$$A