Integral of $$$4 x - 5$$$

Find $$$\int \left(4 x - 5\right)\, dx$$$.

Integrate term by term:

$${\color{red}{\int{\left(4 x - 5\right)d x}}} = {\color{red}{\left(- \int{5 d x} + \int{4 x d x}\right)}}$$

Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=5$$$:

$$\int{4 x d x} - {\color{red}{\int{5 d x}}} = \int{4 x d x} - {\color{red}{\left(5 x\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=4$$$ and $$$f{\left(x \right)} = x$$$:

$$- 5 x + {\color{red}{\int{4 x d x}}} = - 5 x + {\color{red}{\left(4 \int{x d x}\right)}}$$

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

$$- 5 x + 4 {\color{red}{\int{x d x}}}=- 5 x + 4 {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}=- 5 x + 4 {\color{red}{\left(\frac{x^{2}}{2}\right)}}$$

Therefore,

$$\int{\left(4 x - 5\right)d x} = 2 x^{2} - 5 x$$

Simplify:

$$\int{\left(4 x - 5\right)d x} = x \left(2 x - 5\right)$$

Add the constant of integration:

$$\int{\left(4 x - 5\right)d x} = x \left(2 x - 5\right)+C$$

$$$\int \left(4 x - 5\right)\, dx = x \left(2 x - 5\right) + C$$$A