$$$e \left(x - \frac{1}{x}\right)$$$ 的积分

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

Expand the expression:

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

逐项积分：

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

对 $$$c=e$$$ 和 $$$f{\left(x \right)} = x$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$$- \int{\frac{e}{x} d x} + {\color{red}{\int{e x d x}}} = - \int{\frac{e}{x} d x} + {\color{red}{e \int{x d x}}}$$

应用幂法则 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=1$$$：

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

对 $$$c=e$$$ 和 $$$f{\left(x \right)} = \frac{1}{x}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$$\frac{e x^{2}}{2} - {\color{red}{\int{\frac{e}{x} d x}}} = \frac{e x^{2}}{2} - {\color{red}{e \int{\frac{1}{x} d x}}}$$

$$$\frac{1}{x}$$$ 的积分为 $$$\int{\frac{1}{x} d x} = \ln{\left(\left|{x}\right| \right)}$$$:

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

因此，

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

化简：

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

加上积分常数：

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

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