Integral of $$$6 t - 2$$$

Find $$$\int \left(6 t - 2\right)\, dt$$$.

Integrate term by term:

$${\color{red}{\int{\left(6 t - 2\right)d t}}} = {\color{red}{\left(- \int{2 d t} + \int{6 t d t}\right)}}$$

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

$$\int{6 t d t} - {\color{red}{\int{2 d t}}} = \int{6 t d t} - {\color{red}{\left(2 t\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ with $$$c=6$$$ and $$$f{\left(t \right)} = t$$$:

$$- 2 t + {\color{red}{\int{6 t d t}}} = - 2 t + {\color{red}{\left(6 \int{t d t}\right)}}$$

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

$$- 2 t + 6 {\color{red}{\int{t d t}}}=- 2 t + 6 {\color{red}{\frac{t^{1 + 1}}{1 + 1}}}=- 2 t + 6 {\color{red}{\left(\frac{t^{2}}{2}\right)}}$$

Therefore,

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

Simplify:

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

Add the constant of integration:

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

$$$\int \left(6 t - 2\right)\, dt = t \left(3 t - 2\right) + C$$$A