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Integral de $$$- 80 x - \frac{10}{x^{2}}$$$
Calculadora relacionada: Calculadora de integrales definidas e impropias
Tu entrada
Halla $$$\int \left(- 80 x - \frac{10}{x^{2}}\right)\, dx$$$.
Solución
Integra término a término:
$${\color{red}{\int{\left(- 80 x - \frac{10}{x^{2}}\right)d x}}} = {\color{red}{\left(- \int{\frac{10}{x^{2}} d x} - \int{80 x d x}\right)}}$$
Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=80$$$ y $$$f{\left(x \right)} = x$$$:
$$- \int{\frac{10}{x^{2}} d x} - {\color{red}{\int{80 x d x}}} = - \int{\frac{10}{x^{2}} d x} - {\color{red}{\left(80 \int{x d x}\right)}}$$
Aplica la regla de la potencia $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ con $$$n=1$$$:
$$- \int{\frac{10}{x^{2}} d x} - 80 {\color{red}{\int{x d x}}}=- \int{\frac{10}{x^{2}} d x} - 80 {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}=- \int{\frac{10}{x^{2}} d x} - 80 {\color{red}{\left(\frac{x^{2}}{2}\right)}}$$
Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=10$$$ y $$$f{\left(x \right)} = \frac{1}{x^{2}}$$$:
$$- 40 x^{2} - {\color{red}{\int{\frac{10}{x^{2}} d x}}} = - 40 x^{2} - {\color{red}{\left(10 \int{\frac{1}{x^{2}} d x}\right)}}$$
Aplica la regla de la potencia $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ con $$$n=-2$$$:
$$- 40 x^{2} - 10 {\color{red}{\int{\frac{1}{x^{2}} d x}}}=- 40 x^{2} - 10 {\color{red}{\int{x^{-2} d x}}}=- 40 x^{2} - 10 {\color{red}{\frac{x^{-2 + 1}}{-2 + 1}}}=- 40 x^{2} - 10 {\color{red}{\left(- x^{-1}\right)}}=- 40 x^{2} - 10 {\color{red}{\left(- \frac{1}{x}\right)}}$$
Por lo tanto,
$$\int{\left(- 80 x - \frac{10}{x^{2}}\right)d x} = - 40 x^{2} + \frac{10}{x}$$
Simplificar:
$$\int{\left(- 80 x - \frac{10}{x^{2}}\right)d x} = \frac{10 \left(1 - 4 x^{3}\right)}{x}$$
Añade la constante de integración:
$$\int{\left(- 80 x - \frac{10}{x^{2}}\right)d x} = \frac{10 \left(1 - 4 x^{3}\right)}{x}+C$$
Respuesta
$$$\int \left(- 80 x - \frac{10}{x^{2}}\right)\, dx = \frac{10 \left(1 - 4 x^{3}\right)}{x} + C$$$A