$$$\frac{9}{x^{3} \left(3 x - 2\right)}$$$ 的積分

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

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=9$$$ 與 $$$f{\left(x \right)} = \frac{1}{x^{3} \left(3 x - 2\right)}$$$：

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

進行部分分式分解（步驟可見 »）:

$$9 {\color{red}{\int{\frac{1}{x^{3} \left(3 x - 2\right)} d x}}} = 9 {\color{red}{\int{\left(\frac{27}{8 \left(3 x - 2\right)} - \frac{9}{8 x} - \frac{3}{4 x^{2}} - \frac{1}{2 x^{3}}\right)d x}}}$$

逐項積分：

$$9 {\color{red}{\int{\left(\frac{27}{8 \left(3 x - 2\right)} - \frac{9}{8 x} - \frac{3}{4 x^{2}} - \frac{1}{2 x^{3}}\right)d x}}} = 9 {\color{red}{\left(- \int{\frac{1}{2 x^{3}} d x} - \int{\frac{3}{4 x^{2}} d x} - \int{\frac{9}{8 x} d x} + \int{\frac{27}{8 \left(3 x - 2\right)} d x}\right)}}$$

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=\frac{9}{8}$$$ 與 $$$f{\left(x \right)} = \frac{1}{x}$$$：

$$- 9 \int{\frac{1}{2 x^{3}} d x} - 9 \int{\frac{3}{4 x^{2}} d x} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - 9 {\color{red}{\int{\frac{9}{8 x} d x}}} = - 9 \int{\frac{1}{2 x^{3}} d x} - 9 \int{\frac{3}{4 x^{2}} d x} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - 9 {\color{red}{\left(\frac{9 \int{\frac{1}{x} d x}}{8}\right)}}$$

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

$$- 9 \int{\frac{1}{2 x^{3}} d x} - 9 \int{\frac{3}{4 x^{2}} d x} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - \frac{81 {\color{red}{\int{\frac{1}{x} d x}}}}{8} = - 9 \int{\frac{1}{2 x^{3}} d x} - 9 \int{\frac{3}{4 x^{2}} d x} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - \frac{81 {\color{red}{\ln{\left(\left|{x}\right| \right)}}}}{8}$$

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=\frac{3}{4}$$$ 與 $$$f{\left(x \right)} = \frac{1}{x^{2}}$$$：

$$- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} - 9 \int{\frac{1}{2 x^{3}} d x} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - 9 {\color{red}{\int{\frac{3}{4 x^{2}} d x}}} = - \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} - 9 \int{\frac{1}{2 x^{3}} d x} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - 9 {\color{red}{\left(\frac{3 \int{\frac{1}{x^{2}} d x}}{4}\right)}}$$

套用冪次法則 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=-2$$$：

$$- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} - 9 \int{\frac{1}{2 x^{3}} d x} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - \frac{27 {\color{red}{\int{\frac{1}{x^{2}} d x}}}}{4}=- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} - 9 \int{\frac{1}{2 x^{3}} d x} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - \frac{27 {\color{red}{\int{x^{-2} d x}}}}{4}=- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} - 9 \int{\frac{1}{2 x^{3}} d x} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - \frac{27 {\color{red}{\frac{x^{-2 + 1}}{-2 + 1}}}}{4}=- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} - 9 \int{\frac{1}{2 x^{3}} d x} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - \frac{27 {\color{red}{\left(- x^{-1}\right)}}}{4}=- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} - 9 \int{\frac{1}{2 x^{3}} d x} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - \frac{27 {\color{red}{\left(- \frac{1}{x}\right)}}}{4}$$

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

$$- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - 9 {\color{red}{\int{\frac{1}{2 x^{3}} d x}}} + \frac{27}{4 x} = - \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - 9 {\color{red}{\left(\frac{\int{\frac{1}{x^{3}} d x}}{2}\right)}} + \frac{27}{4 x}$$

套用冪次法則 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=-3$$$：

$$- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - \frac{9 {\color{red}{\int{\frac{1}{x^{3}} d x}}}}{2} + \frac{27}{4 x}=- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - \frac{9 {\color{red}{\int{x^{-3} d x}}}}{2} + \frac{27}{4 x}=- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - \frac{9 {\color{red}{\frac{x^{-3 + 1}}{-3 + 1}}}}{2} + \frac{27}{4 x}=- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - \frac{9 {\color{red}{\left(- \frac{x^{-2}}{2}\right)}}}{2} + \frac{27}{4 x}=- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + 9 \int{\frac{27}{8 \left(3 x - 2\right)} d x} - \frac{9 {\color{red}{\left(- \frac{1}{2 x^{2}}\right)}}}{2} + \frac{27}{4 x}$$

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=\frac{27}{8}$$$ 與 $$$f{\left(x \right)} = \frac{1}{3 x - 2}$$$：

$$- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + 9 {\color{red}{\int{\frac{27}{8 \left(3 x - 2\right)} d x}}} + \frac{27}{4 x} + \frac{9}{4 x^{2}} = - \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + 9 {\color{red}{\left(\frac{27 \int{\frac{1}{3 x - 2} d x}}{8}\right)}} + \frac{27}{4 x} + \frac{9}{4 x^{2}}$$

令 $$$u=3 x - 2$$$。

則 $$$du=\left(3 x - 2\right)^{\prime }dx = 3 dx$$$ (步驟見»)，並可得 $$$dx = \frac{du}{3}$$$。

因此，

$$- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + \frac{243 {\color{red}{\int{\frac{1}{3 x - 2} d x}}}}{8} + \frac{27}{4 x} + \frac{9}{4 x^{2}} = - \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + \frac{243 {\color{red}{\int{\frac{1}{3 u} d u}}}}{8} + \frac{27}{4 x} + \frac{9}{4 x^{2}}$$

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=\frac{1}{3}$$$ 與 $$$f{\left(u \right)} = \frac{1}{u}$$$：

$$- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + \frac{243 {\color{red}{\int{\frac{1}{3 u} d u}}}}{8} + \frac{27}{4 x} + \frac{9}{4 x^{2}} = - \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + \frac{243 {\color{red}{\left(\frac{\int{\frac{1}{u} d u}}{3}\right)}}}{8} + \frac{27}{4 x} + \frac{9}{4 x^{2}}$$

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

$$- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + \frac{81 {\color{red}{\int{\frac{1}{u} d u}}}}{8} + \frac{27}{4 x} + \frac{9}{4 x^{2}} = - \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + \frac{81 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{8} + \frac{27}{4 x} + \frac{9}{4 x^{2}}$$

回顧一下 $$$u=3 x - 2$$$：

$$- \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + \frac{81 \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{8} + \frac{27}{4 x} + \frac{9}{4 x^{2}} = - \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + \frac{81 \ln{\left(\left|{{\color{red}{\left(3 x - 2\right)}}}\right| \right)}}{8} + \frac{27}{4 x} + \frac{9}{4 x^{2}}$$

因此，

$$\int{\frac{9}{x^{3} \left(3 x - 2\right)} d x} = - \frac{81 \ln{\left(\left|{x}\right| \right)}}{8} + \frac{81 \ln{\left(\left|{3 x - 2}\right| \right)}}{8} + \frac{27}{4 x} + \frac{9}{4 x^{2}}$$

化簡：

$$\int{\frac{9}{x^{3} \left(3 x - 2\right)} d x} = \frac{9 \left(9 x^{2} \left(- \ln{\left(\left|{x}\right| \right)} + \ln{\left(\left|{3 x - 2}\right| \right)}\right) + 6 x + 2\right)}{8 x^{2}}$$

加上積分常數：

$$\int{\frac{9}{x^{3} \left(3 x - 2\right)} d x} = \frac{9 \left(9 x^{2} \left(- \ln{\left(\left|{x}\right| \right)} + \ln{\left(\left|{3 x - 2}\right| \right)}\right) + 6 x + 2\right)}{8 x^{2}}+C$$

$$$\int \frac{9}{x^{3} \left(3 x - 2\right)}\, dx = \frac{9 \left(9 x^{2} \left(- \ln\left(\left|{x}\right|\right) + \ln\left(\left|{3 x - 2}\right|\right)\right) + 6 x + 2\right)}{8 x^{2}} + C$$$A