Integral of $$$\frac{g}{r^{2}} - e$$$ with respect to $$$g$$$

Find $$$\int \left(\frac{g}{r^{2}} - e\right)\, dg$$$.

Integrate term by term:

$${\color{red}{\int{\left(\frac{g}{r^{2}} - e\right)d g}}} = {\color{red}{\left(- \int{e d g} + \int{\frac{g}{r^{2}} d g}\right)}}$$

Apply the constant rule $$$\int c\, dg = c g$$$ with $$$c=e$$$:

$$\int{\frac{g}{r^{2}} d g} - {\color{red}{\int{e d g}}} = \int{\frac{g}{r^{2}} d g} - {\color{red}{e g}}$$

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

$$- e g + {\color{red}{\int{\frac{g}{r^{2}} d g}}} = - e g + {\color{red}{\frac{\int{g d g}}{r^{2}}}}$$

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

$$- e g + \frac{{\color{red}{\int{g d g}}}}{r^{2}}=- e g + \frac{{\color{red}{\frac{g^{1 + 1}}{1 + 1}}}}{r^{2}}=- e g + \frac{{\color{red}{\left(\frac{g^{2}}{2}\right)}}}{r^{2}}$$

Therefore,

$$\int{\left(\frac{g}{r^{2}} - e\right)d g} = \frac{g^{2}}{2 r^{2}} - e g$$

Add the constant of integration:

$$\int{\left(\frac{g}{r^{2}} - e\right)d g} = \frac{g^{2}}{2 r^{2}} - e g+C$$

$$$\int \left(\frac{g}{r^{2}} - e\right)\, dg = \left(\frac{g^{2}}{2 r^{2}} - e g\right) + C$$$A