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Integral de $$$- a^{2} + \frac{1}{a^{2}}$$$ con respecto a $$$x$$$
Calculadora relacionada: Calculadora de integrales definidas e impropias
Tu entrada
Halla $$$\int \left(- a^{2} + \frac{1}{a^{2}}\right)\, dx$$$.
Solución
Aplica la regla de la constante $$$\int c\, dx = c x$$$ con $$$c=- a^{2} + \frac{1}{a^{2}}$$$:
$${\color{red}{\int{\left(- a^{2} + \frac{1}{a^{2}}\right)d x}}} = {\color{red}{x \left(- a^{2} + \frac{1}{a^{2}}\right)}}$$
Por lo tanto,
$$\int{\left(- a^{2} + \frac{1}{a^{2}}\right)d x} = x \left(- a^{2} + \frac{1}{a^{2}}\right)$$
Simplificar:
$$\int{\left(- a^{2} + \frac{1}{a^{2}}\right)d x} = \frac{x \left(1 - a^{4}\right)}{a^{2}}$$
Añade la constante de integración:
$$\int{\left(- a^{2} + \frac{1}{a^{2}}\right)d x} = \frac{x \left(1 - a^{4}\right)}{a^{2}}+C$$
Respuesta
$$$\int \left(- a^{2} + \frac{1}{a^{2}}\right)\, dx = \frac{x \left(1 - a^{4}\right)}{a^{2}} + C$$$A