$$$\frac{6}{\left(3 x - 2\right)^{3}}$$$ 的積分

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

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=6$$$ 與 $$$f{\left(x \right)} = \frac{1}{\left(3 x - 2\right)^{3}}$$$：

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

令 $$$u=3 x - 2$$$。

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

所以，

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

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

$$6 {\color{red}{\int{\frac{1}{3 u^{3}} d u}}} = 6 {\color{red}{\left(\frac{\int{\frac{1}{u^{3}} d u}}{3}\right)}}$$

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

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

回顧一下 $$$u=3 x - 2$$$：

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

因此，

$$\int{\frac{6}{\left(3 x - 2\right)^{3}} d x} = - \frac{1}{\left(3 x - 2\right)^{2}}$$

加上積分常數：

$$\int{\frac{6}{\left(3 x - 2\right)^{3}} d x} = - \frac{1}{\left(3 x - 2\right)^{2}}+C$$

$$$\int \frac{6}{\left(3 x - 2\right)^{3}}\, dx = - \frac{1}{\left(3 x - 2\right)^{2}} + C$$$A