$$$\frac{1}{x} - \frac{1}{3 x^{3}}$$$ 的积分

求$$$\int \left(\frac{1}{x} - \frac{1}{3 x^{3}}\right)\, dx$$$。

逐项积分：

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

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

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

对 $$$c=\frac{1}{3}$$$ 和 $$$f{\left(x \right)} = \frac{1}{x^{3}}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

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

应用幂法则 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=-3$$$：

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

因此，

$$\int{\left(\frac{1}{x} - \frac{1}{3 x^{3}}\right)d x} = \ln{\left(\left|{x}\right| \right)} + \frac{1}{6 x^{2}}$$

加上积分常数：

$$\int{\left(\frac{1}{x} - \frac{1}{3 x^{3}}\right)d x} = \ln{\left(\left|{x}\right| \right)} + \frac{1}{6 x^{2}}+C$$

$$$\int \left(\frac{1}{x} - \frac{1}{3 x^{3}}\right)\, dx = \left(\ln\left(\left|{x}\right|\right) + \frac{1}{6 x^{2}}\right) + C$$$A