$$$\frac{1}{x^{2} + x - 2}$$$ 的积分

求$$$\int \frac{1}{x^{2} + x - 2}\, dx$$$。

进行部分分式分解（步骤可见»）:

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

逐项积分：

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

对 $$$c=\frac{1}{3}$$$ 和 $$$f{\left(x \right)} = \frac{1}{x + 2}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

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

设$$$u=x + 2$$$。

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

所以，

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

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

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

回忆一下 $$$u=x + 2$$$:

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

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

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

设$$$u=x - 1$$$。

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

积分变为

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

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

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

回忆一下 $$$u=x - 1$$$:

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

因此，

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

加上积分常数：

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

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