$$$\frac{\left(x - 2\right) \left(x - 1\right)}{x}$$$ 的積分

求$$$\int \frac{\left(x - 2\right) \left(x - 1\right)}{x}\, dx$$$。

Expand the expression:

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

逐項積分：

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

配合 $$$c=3$$$，應用常數法則 $$$\int c\, dx = c x$$$：

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

套用冪次法則 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=1$$$：

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

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=2$$$ 與 $$$f{\left(x \right)} = \frac{1}{x}$$$：

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

$$$\frac{1}{x}$$$ 的積分是 $$$\int{\frac{1}{x} d x} = \ln{\left(\left|{x}\right| \right)}$$$：

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

因此，

$$\int{\frac{\left(x - 2\right) \left(x - 1\right)}{x} d x} = \frac{x^{2}}{2} - 3 x + 2 \ln{\left(\left|{x}\right| \right)}$$

加上積分常數：

$$\int{\frac{\left(x - 2\right) \left(x - 1\right)}{x} d x} = \frac{x^{2}}{2} - 3 x + 2 \ln{\left(\left|{x}\right| \right)}+C$$

$$$\int \frac{\left(x - 2\right) \left(x - 1\right)}{x}\, dx = \left(\frac{x^{2}}{2} - 3 x + 2 \ln\left(\left|{x}\right|\right)\right) + C$$$A