$$$\frac{x - 1}{x^{2} + x + 1}$$$ 的積分

求$$$\int \frac{x - 1}{x^{2} + x + 1}\, dx$$$。

將線性項改寫為 $$$x - 1=x\color{red}{+\frac{1}{2}- \frac{1}{2}}-1=x+\frac{1}{2}- \frac{3}{2}$$$，並拆分表達式:

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

逐項積分：

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

令 $$$u=x^{2} + x + 1$$$。

則 $$$du=\left(x^{2} + x + 1\right)^{\prime }dx = \left(2 x + 1\right) dx$$$ (步驟見»)，並可得 $$$\left(2 x + 1\right) dx = du$$$。

該積分可改寫為

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

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=\frac{1}{2}$$$ 與 $$$f{\left(u \right)} = \frac{1}{u}$$$：

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

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

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

回顧一下 $$$u=x^{2} + x + 1$$$：

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

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=- \frac{3}{2}$$$ 與 $$$f{\left(x \right)} = \frac{1}{x^{2} + x + 1}$$$：

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

配方法 (步驟見 »): $$$x^{2} + x + 1 = \left(x + \frac{1}{2}\right)^{2} + \frac{3}{4}$$$:

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

令 $$$u=x + \frac{1}{2}$$$。

則 $$$du=\left(x + \frac{1}{2}\right)^{\prime }dx = 1 dx$$$ (步驟見»)，並可得 $$$dx = du$$$。

所以，

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

令 $$$v=\frac{2 \sqrt{3} u}{3}$$$。

則 $$$dv=\left(\frac{2 \sqrt{3} u}{3}\right)^{\prime }du = \frac{2 \sqrt{3}}{3} du$$$ (步驟見»)，並可得 $$$du = \frac{\sqrt{3} dv}{2}$$$。

該積分可改寫為

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

套用常數倍法則 $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$，使用 $$$c=\frac{2 \sqrt{3}}{3}$$$ 與 $$$f{\left(v \right)} = \frac{1}{v^{2} + 1}$$$：

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

$$$\frac{1}{v^{2} + 1}$$$ 的積分是 $$$\int{\frac{1}{v^{2} + 1} d v} = \operatorname{atan}{\left(v \right)}$$$：

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

回顧一下 $$$v=\frac{2 \sqrt{3} u}{3}$$$：

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

回顧一下 $$$u=x + \frac{1}{2}$$$：

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

因此，

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

化簡：

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

加上積分常數：

$$\int{\frac{x - 1}{x^{2} + x + 1} d x} = \frac{\ln{\left(\left|{x^{2} + x + 1}\right| \right)}}{2} - \sqrt{3} \operatorname{atan}{\left(\frac{\sqrt{3} \left(2 x + 1\right)}{3} \right)}+C$$

$$$\int \frac{x - 1}{x^{2} + x + 1}\, dx = \left(\frac{\ln\left(\left|{x^{2} + x + 1}\right|\right)}{2} - \sqrt{3} \operatorname{atan}{\left(\frac{\sqrt{3} \left(2 x + 1\right)}{3} \right)}\right) + C$$$A