Integral von $$$- \frac{1}{2} + \frac{1}{9 x^{2}}$$$

Bestimme $$$\int \left(- \frac{1}{2} + \frac{1}{9 x^{2}}\right)\, dx$$$.

Gliedweise integrieren:

$${\color{red}{\int{\left(- \frac{1}{2} + \frac{1}{9 x^{2}}\right)d x}}} = {\color{red}{\left(- \int{\frac{1}{2} d x} + \int{\frac{1}{9 x^{2}} d x}\right)}}$$

Wenden Sie die Konstantenregel $$$\int c\, dx = c x$$$ mit $$$c=\frac{1}{2}$$$ an:

$$\int{\frac{1}{9 x^{2}} d x} - {\color{red}{\int{\frac{1}{2} d x}}} = \int{\frac{1}{9 x^{2}} d x} - {\color{red}{\left(\frac{x}{2}\right)}}$$

Wende die Konstantenfaktorregel $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ mit $$$c=\frac{1}{9}$$$ und $$$f{\left(x \right)} = \frac{1}{x^{2}}$$$ an:

$$- \frac{x}{2} + {\color{red}{\int{\frac{1}{9 x^{2}} d x}}} = - \frac{x}{2} + {\color{red}{\left(\frac{\int{\frac{1}{x^{2}} d x}}{9}\right)}}$$

Wenden Sie die Potenzregel $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ mit $$$n=-2$$$ an:

$$- \frac{x}{2} + \frac{{\color{red}{\int{\frac{1}{x^{2}} d x}}}}{9}=- \frac{x}{2} + \frac{{\color{red}{\int{x^{-2} d x}}}}{9}=- \frac{x}{2} + \frac{{\color{red}{\frac{x^{-2 + 1}}{-2 + 1}}}}{9}=- \frac{x}{2} + \frac{{\color{red}{\left(- x^{-1}\right)}}}{9}=- \frac{x}{2} + \frac{{\color{red}{\left(- \frac{1}{x}\right)}}}{9}$$

Daher,

$$\int{\left(- \frac{1}{2} + \frac{1}{9 x^{2}}\right)d x} = - \frac{x}{2} - \frac{1}{9 x}$$

Fügen Sie die Integrationskonstante hinzu:

$$\int{\left(- \frac{1}{2} + \frac{1}{9 x^{2}}\right)d x} = - \frac{x}{2} - \frac{1}{9 x}+C$$

$$$\int \left(- \frac{1}{2} + \frac{1}{9 x^{2}}\right)\, dx = \left(- \frac{x}{2} - \frac{1}{9 x}\right) + C$$$A