Integraal van $$$\sqrt{6} \left(4 x^{7} + 1\right)$$$

Bepaal $$$\int \sqrt{6} \left(4 x^{7} + 1\right)\, dx$$$.

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=\sqrt{6}$$$ en $$$f{\left(x \right)} = 4 x^{7} + 1$$$:

$${\color{red}{\int{\sqrt{6} \left(4 x^{7} + 1\right) d x}}} = {\color{red}{\sqrt{6} \int{\left(4 x^{7} + 1\right)d x}}}$$

Integreer termgewijs:

$$\sqrt{6} {\color{red}{\int{\left(4 x^{7} + 1\right)d x}}} = \sqrt{6} {\color{red}{\left(\int{1 d x} + \int{4 x^{7} d x}\right)}}$$

Pas de constantenregel $$$\int c\, dx = c x$$$ toe met $$$c=1$$$:

$$\sqrt{6} \left(\int{4 x^{7} d x} + {\color{red}{\int{1 d x}}}\right) = \sqrt{6} \left(\int{4 x^{7} d x} + {\color{red}{x}}\right)$$

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=4$$$ en $$$f{\left(x \right)} = x^{7}$$$:

$$\sqrt{6} \left(x + {\color{red}{\int{4 x^{7} d x}}}\right) = \sqrt{6} \left(x + {\color{red}{\left(4 \int{x^{7} d x}\right)}}\right)$$

Pas de machtsregel $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ toe met $$$n=7$$$:

$$\sqrt{6} \left(x + 4 {\color{red}{\int{x^{7} d x}}}\right)=\sqrt{6} \left(x + 4 {\color{red}{\frac{x^{1 + 7}}{1 + 7}}}\right)=\sqrt{6} \left(x + 4 {\color{red}{\left(\frac{x^{8}}{8}\right)}}\right)$$

Dus,

$$\int{\sqrt{6} \left(4 x^{7} + 1\right) d x} = \sqrt{6} \left(\frac{x^{8}}{2} + x\right)$$

Vereenvoudig:

$$\int{\sqrt{6} \left(4 x^{7} + 1\right) d x} = \frac{\sqrt{6} x \left(x^{7} + 2\right)}{2}$$

Voeg de integratieconstante toe:

$$\int{\sqrt{6} \left(4 x^{7} + 1\right) d x} = \frac{\sqrt{6} x \left(x^{7} + 2\right)}{2}+C$$

$$$\int \sqrt{6} \left(4 x^{7} + 1\right)\, dx = \frac{\sqrt{6} x \left(x^{7} + 2\right)}{2} + C$$$A