Integral of $$$- a^{2} + \frac{a}{x^{2}}$$$ with respect to $$$x$$$

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

Integrate term by term:

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

Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=a^{2}$$$:

$$\int{\frac{a}{x^{2}} d x} - {\color{red}{\int{a^{2} d x}}} = \int{\frac{a}{x^{2}} d x} - {\color{red}{a^{2} x}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=a$$$ and $$$f{\left(x \right)} = \frac{1}{x^{2}}$$$:

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

Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=-2$$$:

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

Therefore,

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

Add the constant of integration:

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

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