$$$\frac{2}{\left(x - 1\right)^{\frac{2}{3}}}$$$ 的积分

求$$$\int \frac{2}{\left(x - 1\right)^{\frac{2}{3}}}\, dx$$$。

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

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

设$$$u=x - 1$$$。

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

因此，

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

应用幂法则 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=- \frac{2}{3}$$$：

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

回忆一下 $$$u=x - 1$$$:

$$6 \sqrt[3]{{\color{red}{u}}} = 6 \sqrt[3]{{\color{red}{\left(x - 1\right)}}}$$

因此，

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

加上积分常数：

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

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