$$$\frac{6}{\left(3 x - 2\right)^{3}}$$$ 的积分

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

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

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

设$$$u=3 x - 2$$$。

则$$$du=\left(3 x - 2\right)^{\prime }dx = 3 dx$$$ (步骤见»)，并有$$$dx = \frac{du}{3}$$$。

因此，

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

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

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

应用幂法则 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=-3$$$：

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

回忆一下 $$$u=3 x - 2$$$:

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

因此，

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

加上积分常数：

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

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