$$$\frac{x - 1}{x + 1}$$$ 的积分

求$$$\int \frac{x - 1}{x + 1}\, dx$$$。

设$$$u=x + 1$$$。

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

所以，

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

Expand the expression:

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

逐项积分：

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

应用常数法则 $$$\int c\, du = c u$$$，使用 $$$c=1$$$：

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

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

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

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

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

回忆一下 $$$u=x + 1$$$:

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

因此，

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

加上积分常数（并从表达式中去除常数项）：

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

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