$$$\frac{2 x^{3} - 6 x^{2}}{x - 2}$$$ 的积分

求$$$\int \frac{2 x^{3} - 6 x^{2}}{x - 2}\, dx$$$。

化简被积函数:

$${\color{red}{\int{\frac{2 x^{3} - 6 x^{2}}{x - 2} d x}}} = {\color{red}{\int{\frac{2 x^{2} \left(x - 3\right)}{x - 2} d x}}}$$

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

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

由于分子次数不小于分母次数，进行多项式长除法（步骤见»）:

$$2 {\color{red}{\int{\frac{x^{2} \left(x - 3\right)}{x - 2} d x}}} = 2 {\color{red}{\int{\left(x^{2} - x - 2 - \frac{4}{x - 2}\right)d x}}}$$

逐项积分：

$$2 {\color{red}{\int{\left(x^{2} - x - 2 - \frac{4}{x - 2}\right)d x}}} = 2 {\color{red}{\left(- \int{2 d x} - \int{x d x} + \int{x^{2} d x} - \int{\frac{4}{x - 2} d x}\right)}}$$

应用常数法则 $$$\int c\, dx = c x$$$，使用 $$$c=2$$$：

$$- 2 \int{x d x} + 2 \int{x^{2} d x} - 2 \int{\frac{4}{x - 2} d x} - 2 {\color{red}{\int{2 d x}}} = - 2 \int{x d x} + 2 \int{x^{2} d x} - 2 \int{\frac{4}{x - 2} d x} - 2 {\color{red}{\left(2 x\right)}}$$

应用幂法则 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=2$$$：

$$- 4 x - 2 \int{x d x} - 2 \int{\frac{4}{x - 2} d x} + 2 {\color{red}{\int{x^{2} d x}}}=- 4 x - 2 \int{x d x} - 2 \int{\frac{4}{x - 2} d x} + 2 {\color{red}{\frac{x^{1 + 2}}{1 + 2}}}=- 4 x - 2 \int{x d x} - 2 \int{\frac{4}{x - 2} d x} + 2 {\color{red}{\left(\frac{x^{3}}{3}\right)}}$$

应用幂法则 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=1$$$：

$$\frac{2 x^{3}}{3} - 4 x - 2 \int{\frac{4}{x - 2} d x} - 2 {\color{red}{\int{x d x}}}=\frac{2 x^{3}}{3} - 4 x - 2 \int{\frac{4}{x - 2} d x} - 2 {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}=\frac{2 x^{3}}{3} - 4 x - 2 \int{\frac{4}{x - 2} d x} - 2 {\color{red}{\left(\frac{x^{2}}{2}\right)}}$$

对 $$$c=4$$$ 和 $$$f{\left(x \right)} = \frac{1}{x - 2}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$$\frac{2 x^{3}}{3} - x^{2} - 4 x - 2 {\color{red}{\int{\frac{4}{x - 2} d x}}} = \frac{2 x^{3}}{3} - x^{2} - 4 x - 2 {\color{red}{\left(4 \int{\frac{1}{x - 2} d x}\right)}}$$

设$$$u=x - 2$$$。

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

该积分可以改写为

$$\frac{2 x^{3}}{3} - x^{2} - 4 x - 8 {\color{red}{\int{\frac{1}{x - 2} d x}}} = \frac{2 x^{3}}{3} - x^{2} - 4 x - 8 {\color{red}{\int{\frac{1}{u} d u}}}$$

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

$$\frac{2 x^{3}}{3} - x^{2} - 4 x - 8 {\color{red}{\int{\frac{1}{u} d u}}} = \frac{2 x^{3}}{3} - x^{2} - 4 x - 8 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

回忆一下 $$$u=x - 2$$$:

$$\frac{2 x^{3}}{3} - x^{2} - 4 x - 8 \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = \frac{2 x^{3}}{3} - x^{2} - 4 x - 8 \ln{\left(\left|{{\color{red}{\left(x - 2\right)}}}\right| \right)}$$

因此，

$$\int{\frac{2 x^{3} - 6 x^{2}}{x - 2} d x} = \frac{2 x^{3}}{3} - x^{2} - 4 x - 8 \ln{\left(\left|{x - 2}\right| \right)}$$

加上积分常数：

$$\int{\frac{2 x^{3} - 6 x^{2}}{x - 2} d x} = \frac{2 x^{3}}{3} - x^{2} - 4 x - 8 \ln{\left(\left|{x - 2}\right| \right)}+C$$

$$$\int \frac{2 x^{3} - 6 x^{2}}{x - 2}\, dx = \left(\frac{2 x^{3}}{3} - x^{2} - 4 x - 8 \ln\left(\left|{x - 2}\right|\right)\right) + C$$$A