Integral of $$$\frac{a}{b} + \frac{3}{2}$$$ with respect to $$$a$$$

Find $$$\int \left(\frac{a}{b} + \frac{3}{2}\right)\, da$$$.

Integrate term by term:

$${\color{red}{\int{\left(\frac{a}{b} + \frac{3}{2}\right)d a}}} = {\color{red}{\left(\int{\frac{3}{2} d a} + \int{\frac{a}{b} d a}\right)}}$$

Apply the constant rule $$$\int c\, da = a c$$$ with $$$c=\frac{3}{2}$$$:

$$\int{\frac{a}{b} d a} + {\color{red}{\int{\frac{3}{2} d a}}} = \int{\frac{a}{b} d a} + {\color{red}{\left(\frac{3 a}{2}\right)}}$$

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

$$\frac{3 a}{2} + {\color{red}{\int{\frac{a}{b} d a}}} = \frac{3 a}{2} + {\color{red}{\frac{\int{a d a}}{b}}}$$

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

$$\frac{3 a}{2} + \frac{{\color{red}{\int{a d a}}}}{b}=\frac{3 a}{2} + \frac{{\color{red}{\frac{a^{1 + 1}}{1 + 1}}}}{b}=\frac{3 a}{2} + \frac{{\color{red}{\left(\frac{a^{2}}{2}\right)}}}{b}$$

Therefore,

$$\int{\left(\frac{a}{b} + \frac{3}{2}\right)d a} = \frac{a^{2}}{2 b} + \frac{3 a}{2}$$

Simplify:

$$\int{\left(\frac{a}{b} + \frac{3}{2}\right)d a} = \frac{a \left(a + 3 b\right)}{2 b}$$

Add the constant of integration:

$$\int{\left(\frac{a}{b} + \frac{3}{2}\right)d a} = \frac{a \left(a + 3 b\right)}{2 b}+C$$

$$$\int \left(\frac{a}{b} + \frac{3}{2}\right)\, da = \frac{a \left(a + 3 b\right)}{2 b} + C$$$A