Integralen av $$$- \frac{1}{x^{3}}$$$

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

Tillämpa konstantfaktorregeln $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ med $$$c=-1$$$ och $$$f{\left(x \right)} = \frac{1}{x^{3}}$$$:

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

Tillämpa potensregeln $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ med $$$n=-3$$$:

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

Alltså,

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

Lägg till integrationskonstanten:

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

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