$$$\frac{x}{\left(x + 1\right) \left(x + 2\right)}$$$ 的积分

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

进行部分分式分解（步骤可见»）:

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

逐项积分：

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

设$$$u=x + 1$$$。

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

因此，

$$\int{\frac{2}{x + 2} d x} - {\color{red}{\int{\frac{1}{x + 1} d x}}} = \int{\frac{2}{x + 2} d x} - {\color{red}{\int{\frac{1}{u} d u}}}$$

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

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

回忆一下 $$$u=x + 1$$$:

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

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

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

设$$$u=x + 2$$$。

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

因此，

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

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

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

回忆一下 $$$u=x + 2$$$:

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

因此，

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

加上积分常数：

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

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