$$$\frac{1}{\left(3 - 5 x\right)^{2}}$$$ 的积分

求$$$\int \frac{1}{\left(3 - 5 x\right)^{2}}\, dx$$$。

设$$$u=3 - 5 x$$$。

则$$$du=\left(3 - 5 x\right)^{\prime }dx = - 5 dx$$$ (步骤见»)，并有$$$dx = - \frac{du}{5}$$$。

积分变为

$${\color{red}{\int{\frac{1}{\left(3 - 5 x\right)^{2}} d x}}} = {\color{red}{\int{\left(- \frac{1}{5 u^{2}}\right)d u}}}$$

对 $$$c=- \frac{1}{5}$$$ 和 $$$f{\left(u \right)} = \frac{1}{u^{2}}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

$${\color{red}{\int{\left(- \frac{1}{5 u^{2}}\right)d u}}} = {\color{red}{\left(- \frac{\int{\frac{1}{u^{2}} d u}}{5}\right)}}$$

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

$$- \frac{{\color{red}{\int{\frac{1}{u^{2}} d u}}}}{5}=- \frac{{\color{red}{\int{u^{-2} d u}}}}{5}=- \frac{{\color{red}{\frac{u^{-2 + 1}}{-2 + 1}}}}{5}=- \frac{{\color{red}{\left(- u^{-1}\right)}}}{5}=- \frac{{\color{red}{\left(- \frac{1}{u}\right)}}}{5}$$

回忆一下 $$$u=3 - 5 x$$$:

$$\frac{{\color{red}{u}}^{-1}}{5} = \frac{{\color{red}{\left(3 - 5 x\right)}}^{-1}}{5}$$

因此，

$$\int{\frac{1}{\left(3 - 5 x\right)^{2}} d x} = \frac{1}{5 \left(3 - 5 x\right)}$$

化简：

$$\int{\frac{1}{\left(3 - 5 x\right)^{2}} d x} = - \frac{1}{25 x - 15}$$

加上积分常数：

$$\int{\frac{1}{\left(3 - 5 x\right)^{2}} d x} = - \frac{1}{25 x - 15}+C$$

$$$\int \frac{1}{\left(3 - 5 x\right)^{2}}\, dx = - \frac{1}{25 x - 15} + C$$$A