Integral of $$$- a + \frac{1}{b}$$$ with respect to $$$a$$$

Find $$$\int \left(- a + \frac{1}{b}\right)\, da$$$.

Integrate term by term:

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

Apply the constant rule $$$\int c\, da = a c$$$ with $$$c=\frac{1}{b}$$$:

$$- \int{a d a} + {\color{red}{\int{\frac{1}{b} d a}}} = - \int{a d a} + {\color{red}{\frac{a}{b}}}$$

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

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

Therefore,

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

Add the constant of integration:

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

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