Integral von $$$-1 + \frac{1}{x^{3}}$$$

Bestimme $$$\int \left(-1 + \frac{1}{x^{3}}\right)\, dx$$$.

Gliedweise integrieren:

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

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

$$\int{\frac{1}{x^{3}} d x} - {\color{red}{\int{1 d x}}} = \int{\frac{1}{x^{3}} d x} - {\color{red}{x}}$$

Wenden Sie die Potenzregel $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ mit $$$n=-3$$$ an:

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

Daher,

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

Fügen Sie die Integrationskonstante hinzu:

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

$$$\int \left(-1 + \frac{1}{x^{3}}\right)\, dx = \left(- x - \frac{1}{2 x^{2}}\right) + C$$$A