Integral de $$$- a^{2} + \frac{1}{a^{2}}$$$ em relação a $$$x$$$

Encontre $$$\int \left(- a^{2} + \frac{1}{a^{2}}\right)\, dx$$$.

Aplique a regra da constante $$$\int c\, dx = c x$$$ usando $$$c=- a^{2} + \frac{1}{a^{2}}$$$:

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

Portanto,

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

Simplifique:

$$\int{\left(- a^{2} + \frac{1}{a^{2}}\right)d x} = \frac{x \left(1 - a^{4}\right)}{a^{2}}$$

Adicione a constante de integração:

$$\int{\left(- a^{2} + \frac{1}{a^{2}}\right)d x} = \frac{x \left(1 - a^{4}\right)}{a^{2}}+C$$

$$$\int \left(- a^{2} + \frac{1}{a^{2}}\right)\, dx = \frac{x \left(1 - a^{4}\right)}{a^{2}} + C$$$A