$$$\frac{1}{x^{3} + x}$$$ 的积分

求$$$\int \frac{1}{x^{3} + x}\, dx$$$。

进行部分分式分解（步骤可见»）:

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

逐项积分：

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

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

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

设$$$u=x^{2} + 1$$$。

则$$$du=\left(x^{2} + 1\right)^{\prime }dx = 2 x dx$$$ (步骤见»)，并有$$$x dx = \frac{du}{2}$$$。

因此，

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

对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(u \right)} = \frac{1}{u}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

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

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

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

回忆一下 $$$u=x^{2} + 1$$$:

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

因此，

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

加上积分常数：

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

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