Integral von $$$\frac{1}{2 x^{4}}$$$

Bestimme $$$\int \frac{1}{2 x^{4}}\, dx$$$.

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=\frac{1}{2}$$$ und $$$f{\left(x \right)} = \frac{1}{x^{4}}$$$ an:

$${\color{red}{\int{\frac{1}{2 x^{4}} d x}}} = {\color{red}{\left(\frac{\int{\frac{1}{x^{4}} d x}}{2}\right)}}$$

Wenden Sie die Potenzregel $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ mit $$$n=-4$$$ an:

$$\frac{{\color{red}{\int{\frac{1}{x^{4}} d x}}}}{2}=\frac{{\color{red}{\int{x^{-4} d x}}}}{2}=\frac{{\color{red}{\frac{x^{-4 + 1}}{-4 + 1}}}}{2}=\frac{{\color{red}{\left(- \frac{x^{-3}}{3}\right)}}}{2}=\frac{{\color{red}{\left(- \frac{1}{3 x^{3}}\right)}}}{2}$$

Daher,

$$\int{\frac{1}{2 x^{4}} d x} = - \frac{1}{6 x^{3}}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\frac{1}{2 x^{4}} d x} = - \frac{1}{6 x^{3}}+C$$

$$$\int \frac{1}{2 x^{4}}\, dx = - \frac{1}{6 x^{3}} + C$$$A