$$$\frac{x + 3}{x - 3}$$$ 的積分

求$$$\int \frac{x + 3}{x - 3}\, dx$$$。

令 $$$u=x - 3$$$。

則 $$$du=\left(x - 3\right)^{\prime }dx = 1 dx$$$ (步驟見»)，並可得 $$$dx = du$$$。

該積分變為

$${\color{red}{\int{\frac{x + 3}{x - 3} d x}}} = {\color{red}{\int{\frac{u + 6}{u} d u}}}$$

Expand the expression:

$${\color{red}{\int{\frac{u + 6}{u} d u}}} = {\color{red}{\int{\left(1 + \frac{6}{u}\right)d u}}}$$

逐項積分：

$${\color{red}{\int{\left(1 + \frac{6}{u}\right)d u}}} = {\color{red}{\left(\int{1 d u} + \int{\frac{6}{u} d u}\right)}}$$

配合 $$$c=1$$$，應用常數法則 $$$\int c\, du = c u$$$：

$$\int{\frac{6}{u} d u} + {\color{red}{\int{1 d u}}} = \int{\frac{6}{u} d u} + {\color{red}{u}}$$

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=6$$$ 與 $$$f{\left(u \right)} = \frac{1}{u}$$$：

$$u + {\color{red}{\int{\frac{6}{u} d u}}} = u + {\color{red}{\left(6 \int{\frac{1}{u} d u}\right)}}$$

$$$\frac{1}{u}$$$ 的積分是 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$：

$$u + 6 {\color{red}{\int{\frac{1}{u} d u}}} = u + 6 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

回顧一下 $$$u=x - 3$$$：

$$6 \ln{\left(\left|{{\color{red}{u}}}\right| \right)} + {\color{red}{u}} = 6 \ln{\left(\left|{{\color{red}{\left(x - 3\right)}}}\right| \right)} + {\color{red}{\left(x - 3\right)}}$$

因此，

$$\int{\frac{x + 3}{x - 3} d x} = x + 6 \ln{\left(\left|{x - 3}\right| \right)} - 3$$

加上積分常數（並從表達式中移除常數項）：

$$\int{\frac{x + 3}{x - 3} d x} = x + 6 \ln{\left(\left|{x - 3}\right| \right)}+C$$

$$$\int \frac{x + 3}{x - 3}\, dx = \left(x + 6 \ln\left(\left|{x - 3}\right|\right)\right) + C$$$A