$$$\frac{x^{4}}{x^{4} - 9}$$$ 的积分

求$$$\int \frac{x^{4}}{x^{4} - 9}\, dx$$$。

改写并拆分该分式:

$${\color{red}{\int{\frac{x^{4}}{x^{4} - 9} d x}}} = {\color{red}{\int{\left(1 + \frac{9}{x^{4} - 9}\right)d x}}}$$

逐项积分：

$${\color{red}{\int{\left(1 + \frac{9}{x^{4} - 9}\right)d x}}} = {\color{red}{\left(\int{1 d x} + \int{\frac{9}{x^{4} - 9} d x}\right)}}$$

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

$$\int{\frac{9}{x^{4} - 9} d x} + {\color{red}{\int{1 d x}}} = \int{\frac{9}{x^{4} - 9} d x} + {\color{red}{x}}$$

对 $$$c=9$$$ 和 $$$f{\left(x \right)} = \frac{1}{x^{4} - 9}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$$x + {\color{red}{\int{\frac{9}{x^{4} - 9} d x}}} = x + {\color{red}{\left(9 \int{\frac{1}{x^{4} - 9} d x}\right)}}$$

进行部分分式分解（步骤可见»）:

$$x + 9 {\color{red}{\int{\frac{1}{x^{4} - 9} d x}}} = x + 9 {\color{red}{\int{\left(- \frac{1}{6 \left(x^{2} + 3\right)} - \frac{\sqrt{3}}{36 \left(x + \sqrt{3}\right)} + \frac{\sqrt{3}}{36 \left(x - \sqrt{3}\right)}\right)d x}}}$$

逐项积分：

$$x + 9 {\color{red}{\int{\left(- \frac{1}{6 \left(x^{2} + 3\right)} - \frac{\sqrt{3}}{36 \left(x + \sqrt{3}\right)} + \frac{\sqrt{3}}{36 \left(x - \sqrt{3}\right)}\right)d x}}} = x + 9 {\color{red}{\left(\int{\frac{\sqrt{3}}{36 \left(x - \sqrt{3}\right)} d x} - \int{\frac{\sqrt{3}}{36 \left(x + \sqrt{3}\right)} d x} - \int{\frac{1}{6 \left(x^{2} + 3\right)} d x}\right)}}$$

对 $$$c=\frac{1}{6}$$$ 和 $$$f{\left(x \right)} = \frac{1}{x^{2} + 3}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$$x + 9 \int{\frac{\sqrt{3}}{36 \left(x - \sqrt{3}\right)} d x} - 9 \int{\frac{\sqrt{3}}{36 \left(x + \sqrt{3}\right)} d x} - 9 {\color{red}{\int{\frac{1}{6 \left(x^{2} + 3\right)} d x}}} = x + 9 \int{\frac{\sqrt{3}}{36 \left(x - \sqrt{3}\right)} d x} - 9 \int{\frac{\sqrt{3}}{36 \left(x + \sqrt{3}\right)} d x} - 9 {\color{red}{\left(\frac{\int{\frac{1}{x^{2} + 3} d x}}{6}\right)}}$$

设$$$u=\frac{\sqrt{3}}{3} x$$$。

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

因此，

$$x + 9 \int{\frac{\sqrt{3}}{36 \left(x - \sqrt{3}\right)} d x} - 9 \int{\frac{\sqrt{3}}{36 \left(x + \sqrt{3}\right)} d x} - \frac{3 {\color{red}{\int{\frac{1}{x^{2} + 3} d x}}}}{2} = x + 9 \int{\frac{\sqrt{3}}{36 \left(x - \sqrt{3}\right)} d x} - 9 \int{\frac{\sqrt{3}}{36 \left(x + \sqrt{3}\right)} d x} - \frac{3 {\color{red}{\int{\frac{\sqrt{3}}{3 \left(u^{2} + 1\right)} d u}}}}{2}$$

对 $$$c=\frac{\sqrt{3}}{3}$$$ 和 $$$f{\left(u \right)} = \frac{1}{u^{2} + 1}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

$$x + 9 \int{\frac{\sqrt{3}}{36 \left(x - \sqrt{3}\right)} d x} - 9 \int{\frac{\sqrt{3}}{36 \left(x + \sqrt{3}\right)} d x} - \frac{3 {\color{red}{\int{\frac{\sqrt{3}}{3 \left(u^{2} + 1\right)} d u}}}}{2} = x + 9 \int{\frac{\sqrt{3}}{36 \left(x - \sqrt{3}\right)} d x} - 9 \int{\frac{\sqrt{3}}{36 \left(x + \sqrt{3}\right)} d x} - \frac{3 {\color{red}{\left(\frac{\sqrt{3} \int{\frac{1}{u^{2} + 1} d u}}{3}\right)}}}{2}$$

$$$\frac{1}{u^{2} + 1}$$$ 的积分为 $$$\int{\frac{1}{u^{2} + 1} d u} = \operatorname{atan}{\left(u \right)}$$$:

$$x + 9 \int{\frac{\sqrt{3}}{36 \left(x - \sqrt{3}\right)} d x} - 9 \int{\frac{\sqrt{3}}{36 \left(x + \sqrt{3}\right)} d x} - \frac{\sqrt{3} {\color{red}{\int{\frac{1}{u^{2} + 1} d u}}}}{2} = x + 9 \int{\frac{\sqrt{3}}{36 \left(x - \sqrt{3}\right)} d x} - 9 \int{\frac{\sqrt{3}}{36 \left(x + \sqrt{3}\right)} d x} - \frac{\sqrt{3} {\color{red}{\operatorname{atan}{\left(u \right)}}}}{2}$$

回忆一下 $$$u=\frac{\sqrt{3}}{3} x$$$:

$$x + 9 \int{\frac{\sqrt{3}}{36 \left(x - \sqrt{3}\right)} d x} - 9 \int{\frac{\sqrt{3}}{36 \left(x + \sqrt{3}\right)} d x} - \frac{\sqrt{3} \operatorname{atan}{\left({\color{red}{u}} \right)}}{2} = x + 9 \int{\frac{\sqrt{3}}{36 \left(x - \sqrt{3}\right)} d x} - 9 \int{\frac{\sqrt{3}}{36 \left(x + \sqrt{3}\right)} d x} - \frac{\sqrt{3} \operatorname{atan}{\left({\color{red}{\frac{\sqrt{3}}{3} x}} \right)}}{2}$$

对 $$$c=\frac{\sqrt{3}}{36}$$$ 和 $$$f{\left(x \right)} = \frac{1}{x + \sqrt{3}}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$$x - \frac{\sqrt{3} \operatorname{atan}{\left(\frac{\sqrt{3}}{3} x \right)}}{2} + 9 \int{\frac{\sqrt{3}}{36 \left(x - \sqrt{3}\right)} d x} - 9 {\color{red}{\int{\frac{\sqrt{3}}{36 \left(x + \sqrt{3}\right)} d x}}} = x - \frac{\sqrt{3} \operatorname{atan}{\left(\frac{\sqrt{3}}{3} x \right)}}{2} + 9 \int{\frac{\sqrt{3}}{36 \left(x - \sqrt{3}\right)} d x} - 9 {\color{red}{\left(\frac{\sqrt{3} \int{\frac{1}{x + \sqrt{3}} d x}}{36}\right)}}$$

设$$$u=x + \sqrt{3}$$$。

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

所以，

$$x - \frac{\sqrt{3} \operatorname{atan}{\left(\frac{\sqrt{3}}{3} x \right)}}{2} + 9 \int{\frac{\sqrt{3}}{36 \left(x - \sqrt{3}\right)} d x} - \frac{\sqrt{3} {\color{red}{\int{\frac{1}{x + \sqrt{3}} d x}}}}{4} = x - \frac{\sqrt{3} \operatorname{atan}{\left(\frac{\sqrt{3}}{3} x \right)}}{2} + 9 \int{\frac{\sqrt{3}}{36 \left(x - \sqrt{3}\right)} d x} - \frac{\sqrt{3} {\color{red}{\int{\frac{1}{u} d u}}}}{4}$$

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

$$x - \frac{\sqrt{3} \operatorname{atan}{\left(\frac{\sqrt{3}}{3} x \right)}}{2} + 9 \int{\frac{\sqrt{3}}{36 \left(x - \sqrt{3}\right)} d x} - \frac{\sqrt{3} {\color{red}{\int{\frac{1}{u} d u}}}}{4} = x - \frac{\sqrt{3} \operatorname{atan}{\left(\frac{\sqrt{3}}{3} x \right)}}{2} + 9 \int{\frac{\sqrt{3}}{36 \left(x - \sqrt{3}\right)} d x} - \frac{\sqrt{3} {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{4}$$

回忆一下 $$$u=x + \sqrt{3}$$$:

$$x - \frac{\sqrt{3} \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{4} - \frac{\sqrt{3} \operatorname{atan}{\left(\frac{\sqrt{3}}{3} x \right)}}{2} + 9 \int{\frac{\sqrt{3}}{36 \left(x - \sqrt{3}\right)} d x} = x - \frac{\sqrt{3} \ln{\left(\left|{{\color{red}{\left(x + \sqrt{3}\right)}}}\right| \right)}}{4} - \frac{\sqrt{3} \operatorname{atan}{\left(\frac{\sqrt{3}}{3} x \right)}}{2} + 9 \int{\frac{\sqrt{3}}{36 \left(x - \sqrt{3}\right)} d x}$$

对 $$$c=\frac{\sqrt{3}}{36}$$$ 和 $$$f{\left(x \right)} = \frac{1}{x - \sqrt{3}}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$$x - \frac{\sqrt{3} \ln{\left(\left|{x + \sqrt{3}}\right| \right)}}{4} - \frac{\sqrt{3} \operatorname{atan}{\left(\frac{\sqrt{3}}{3} x \right)}}{2} + 9 {\color{red}{\int{\frac{\sqrt{3}}{36 \left(x - \sqrt{3}\right)} d x}}} = x - \frac{\sqrt{3} \ln{\left(\left|{x + \sqrt{3}}\right| \right)}}{4} - \frac{\sqrt{3} \operatorname{atan}{\left(\frac{\sqrt{3}}{3} x \right)}}{2} + 9 {\color{red}{\left(\frac{\sqrt{3} \int{\frac{1}{x - \sqrt{3}} d x}}{36}\right)}}$$

设$$$u=x - \sqrt{3}$$$。

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

该积分可以改写为

$$x - \frac{\sqrt{3} \ln{\left(\left|{x + \sqrt{3}}\right| \right)}}{4} - \frac{\sqrt{3} \operatorname{atan}{\left(\frac{\sqrt{3}}{3} x \right)}}{2} + \frac{\sqrt{3} {\color{red}{\int{\frac{1}{x - \sqrt{3}} d x}}}}{4} = x - \frac{\sqrt{3} \ln{\left(\left|{x + \sqrt{3}}\right| \right)}}{4} - \frac{\sqrt{3} \operatorname{atan}{\left(\frac{\sqrt{3}}{3} x \right)}}{2} + \frac{\sqrt{3} {\color{red}{\int{\frac{1}{u} d u}}}}{4}$$

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

$$x - \frac{\sqrt{3} \ln{\left(\left|{x + \sqrt{3}}\right| \right)}}{4} - \frac{\sqrt{3} \operatorname{atan}{\left(\frac{\sqrt{3}}{3} x \right)}}{2} + \frac{\sqrt{3} {\color{red}{\int{\frac{1}{u} d u}}}}{4} = x - \frac{\sqrt{3} \ln{\left(\left|{x + \sqrt{3}}\right| \right)}}{4} - \frac{\sqrt{3} \operatorname{atan}{\left(\frac{\sqrt{3}}{3} x \right)}}{2} + \frac{\sqrt{3} {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{4}$$

回忆一下 $$$u=x - \sqrt{3}$$$:

$$x - \frac{\sqrt{3} \ln{\left(\left|{x + \sqrt{3}}\right| \right)}}{4} + \frac{\sqrt{3} \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{4} - \frac{\sqrt{3} \operatorname{atan}{\left(\frac{\sqrt{3}}{3} x \right)}}{2} = x - \frac{\sqrt{3} \ln{\left(\left|{x + \sqrt{3}}\right| \right)}}{4} + \frac{\sqrt{3} \ln{\left(\left|{{\color{red}{\left(x - \sqrt{3}\right)}}}\right| \right)}}{4} - \frac{\sqrt{3} \operatorname{atan}{\left(\frac{\sqrt{3}}{3} x \right)}}{2}$$

因此，

$$\int{\frac{x^{4}}{x^{4} - 9} d x} = x + \frac{\sqrt{3} \ln{\left(\left|{x - \sqrt{3}}\right| \right)}}{4} - \frac{\sqrt{3} \ln{\left(\left|{x + \sqrt{3}}\right| \right)}}{4} - \frac{\sqrt{3} \operatorname{atan}{\left(\frac{\sqrt{3} x}{3} \right)}}{2}$$

加上积分常数：

$$\int{\frac{x^{4}}{x^{4} - 9} d x} = x + \frac{\sqrt{3} \ln{\left(\left|{x - \sqrt{3}}\right| \right)}}{4} - \frac{\sqrt{3} \ln{\left(\left|{x + \sqrt{3}}\right| \right)}}{4} - \frac{\sqrt{3} \operatorname{atan}{\left(\frac{\sqrt{3} x}{3} \right)}}{2}+C$$

$$$\int \frac{x^{4}}{x^{4} - 9}\, dx = \left(x + \frac{\sqrt{3} \ln\left(\left|{x - \sqrt{3}}\right|\right)}{4} - \frac{\sqrt{3} \ln\left(\left|{x + \sqrt{3}}\right|\right)}{4} - \frac{\sqrt{3} \operatorname{atan}{\left(\frac{\sqrt{3} x}{3} \right)}}{2}\right) + C$$$A