Integral of $$$a^{2} - 3$$$

Find $$$\int \left(a^{2} - 3\right)\, da$$$.

Integrate term by term:

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

Apply the constant rule $$$\int c\, da = a c$$$ with $$$c=3$$$:

$$\int{a^{2} d a} - {\color{red}{\int{3 d a}}} = \int{a^{2} d a} - {\color{red}{\left(3 a\right)}}$$

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

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

Therefore,

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

Simplify:

$$\int{\left(a^{2} - 3\right)d a} = \frac{a \left(a^{2} - 9\right)}{3}$$

Add the constant of integration:

$$\int{\left(a^{2} - 3\right)d a} = \frac{a \left(a^{2} - 9\right)}{3}+C$$

$$$\int \left(a^{2} - 3\right)\, da = \frac{a \left(a^{2} - 9\right)}{3} + C$$$A