$$$\frac{2 x^{3}}{x^{2} - 9}$$$ 的积分

求$$$\int \frac{2 x^{3}}{x^{2} - 9}\, dx$$$。

对 $$$c=2$$$ 和 $$$f{\left(x \right)} = \frac{x^{3}}{x^{2} - 9}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$${\color{red}{\int{\frac{2 x^{3}}{x^{2} - 9} d x}}} = {\color{red}{\left(2 \int{\frac{x^{3}}{x^{2} - 9} d x}\right)}}$$

由于分子次数不小于分母次数，进行多项式长除法（步骤见»）:

$$2 {\color{red}{\int{\frac{x^{3}}{x^{2} - 9} d x}}} = 2 {\color{red}{\int{\left(x + \frac{9 x}{x^{2} - 9}\right)d x}}}$$

逐项积分：

$$2 {\color{red}{\int{\left(x + \frac{9 x}{x^{2} - 9}\right)d x}}} = 2 {\color{red}{\left(\int{x d x} + \int{\frac{9 x}{x^{2} - 9} d x}\right)}}$$

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

$$2 \int{\frac{9 x}{x^{2} - 9} d x} + 2 {\color{red}{\int{x d x}}}=2 \int{\frac{9 x}{x^{2} - 9} d x} + 2 {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}=2 \int{\frac{9 x}{x^{2} - 9} d x} + 2 {\color{red}{\left(\frac{x^{2}}{2}\right)}}$$

设$$$u=x^{2} - 9$$$。

则$$$du=\left(x^{2} - 9\right)^{\prime }dx = 2 x dx$$$ (步骤见»)，并有$$$x dx = \frac{du}{2}$$$。

因此，

$$x^{2} + 2 {\color{red}{\int{\frac{9 x}{x^{2} - 9} d x}}} = x^{2} + 2 {\color{red}{\int{\frac{9}{2 u} d u}}}$$

对 $$$c=\frac{9}{2}$$$ 和 $$$f{\left(u \right)} = \frac{1}{u}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

$$x^{2} + 2 {\color{red}{\int{\frac{9}{2 u} d u}}} = x^{2} + 2 {\color{red}{\left(\frac{9 \int{\frac{1}{u} d u}}{2}\right)}}$$

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

$$x^{2} + 9 {\color{red}{\int{\frac{1}{u} d u}}} = x^{2} + 9 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

回忆一下 $$$u=x^{2} - 9$$$:

$$x^{2} + 9 \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = x^{2} + 9 \ln{\left(\left|{{\color{red}{\left(x^{2} - 9\right)}}}\right| \right)}$$

因此，

$$\int{\frac{2 x^{3}}{x^{2} - 9} d x} = x^{2} + 9 \ln{\left(\left|{x^{2} - 9}\right| \right)}$$

加上积分常数：

$$\int{\frac{2 x^{3}}{x^{2} - 9} d x} = x^{2} + 9 \ln{\left(\left|{x^{2} - 9}\right| \right)}+C$$

$$$\int \frac{2 x^{3}}{x^{2} - 9}\, dx = \left(x^{2} + 9 \ln\left(\left|{x^{2} - 9}\right|\right)\right) + C$$$A