Integral de $$$1 + \frac{1}{x^{9}}$$$

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

Integra término a término:

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

Aplica la regla de la constante $$$\int c\, dx = c x$$$ con $$$c=1$$$:

$$\int{\frac{1}{x^{9}} d x} + {\color{red}{\int{1 d x}}} = \int{\frac{1}{x^{9}} d x} + {\color{red}{x}}$$

Aplica la regla de la potencia $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ con $$$n=-9$$$:

$$x + {\color{red}{\int{\frac{1}{x^{9}} d x}}}=x + {\color{red}{\int{x^{-9} d x}}}=x + {\color{red}{\frac{x^{-9 + 1}}{-9 + 1}}}=x + {\color{red}{\left(- \frac{x^{-8}}{8}\right)}}=x + {\color{red}{\left(- \frac{1}{8 x^{8}}\right)}}$$

Por lo tanto,

$$\int{\left(1 + \frac{1}{x^{9}}\right)d x} = x - \frac{1}{8 x^{8}}$$

Añade la constante de integración:

$$\int{\left(1 + \frac{1}{x^{9}}\right)d x} = x - \frac{1}{8 x^{8}}+C$$

$$$\int \left(1 + \frac{1}{x^{9}}\right)\, dx = \left(x - \frac{1}{8 x^{8}}\right) + C$$$A