Integral of $$$\sqrt{3} t^{2} - 2$$$

Find $$$\int \left(\sqrt{3} t^{2} - 2\right)\, dt$$$.

Integrate term by term:

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

Apply the constant rule $$$\int c\, dt = c t$$$ with $$$c=2$$$:

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

Apply the constant multiple rule $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ with $$$c=\sqrt{3}$$$ and $$$f{\left(t \right)} = t^{2}$$$:

$$- 2 t + {\color{red}{\int{\sqrt{3} t^{2} d t}}} = - 2 t + {\color{red}{\sqrt{3} \int{t^{2} d t}}}$$

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

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

Therefore,

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

Simplify:

$$\int{\left(\sqrt{3} t^{2} - 2\right)d t} = \frac{t \left(\sqrt{3} t^{2} - 6\right)}{3}$$

Add the constant of integration:

$$\int{\left(\sqrt{3} t^{2} - 2\right)d t} = \frac{t \left(\sqrt{3} t^{2} - 6\right)}{3}+C$$

$$$\int \left(\sqrt{3} t^{2} - 2\right)\, dt = \frac{t \left(\sqrt{3} t^{2} - 6\right)}{3} + C$$$A