Integral de $$$- \frac{1}{x^{2}}$$$

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

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=-1$$$ y $$$f{\left(x \right)} = \frac{1}{x^{2}}$$$:

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

Aplica la regla de la potencia $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ con $$$n=-2$$$:

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

Por lo tanto,

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

Añade la constante de integración:

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

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