Integraal van $$$\frac{1}{8 x^{9}}$$$

Bepaal $$$\int \frac{1}{8 x^{9}}\, dx$$$.

Pas de constante-veelvoudregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ toe met $$$c=\frac{1}{8}$$$ en $$$f{\left(x \right)} = \frac{1}{x^{9}}$$$:

$${\color{red}{\int{\frac{1}{8 x^{9}} d x}}} = {\color{red}{\left(\frac{\int{\frac{1}{x^{9}} d x}}{8}\right)}}$$

Pas de machtsregel $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ toe met $$$n=-9$$$:

$$\frac{{\color{red}{\int{\frac{1}{x^{9}} d x}}}}{8}=\frac{{\color{red}{\int{x^{-9} d x}}}}{8}=\frac{{\color{red}{\frac{x^{-9 + 1}}{-9 + 1}}}}{8}=\frac{{\color{red}{\left(- \frac{x^{-8}}{8}\right)}}}{8}=\frac{{\color{red}{\left(- \frac{1}{8 x^{8}}\right)}}}{8}$$

Dus,

$$\int{\frac{1}{8 x^{9}} d x} = - \frac{1}{64 x^{8}}$$

Voeg de integratieconstante toe:

$$\int{\frac{1}{8 x^{9}} d x} = - \frac{1}{64 x^{8}}+C$$

$$$\int \frac{1}{8 x^{9}}\, dx = - \frac{1}{64 x^{8}} + C$$$A