$$$\frac{x}{x + 3}$$$ 的积分

求$$$\int \frac{x}{x + 3}\, dx$$$。

改写并拆分该分式:

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

逐项积分：

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

应用常数法则 $$$\int c\, dx = c x$$$，使用 $$$c=1$$$：

$$- \int{\frac{3}{x + 3} d x} + {\color{red}{\int{1 d x}}} = - \int{\frac{3}{x + 3} d x} + {\color{red}{x}}$$

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

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

设$$$u=x + 3$$$。

则$$$du=\left(x + 3\right)^{\prime }dx = 1 dx$$$ (步骤见»)，并有$$$dx = du$$$。

所以，

$$x - 3 {\color{red}{\int{\frac{1}{x + 3} d x}}} = x - 3 {\color{red}{\int{\frac{1}{u} d u}}}$$

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

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

回忆一下 $$$u=x + 3$$$:

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

因此，

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

加上积分常数：

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

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