$$$- \alpha \beta - x + \frac{1}{x}$$$ 對 $$$x$$$ 的積分

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

逐項積分：

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

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

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

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

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

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

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

因此，

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

加上積分常數：

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

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