$$$\frac{s \sec^{2}{\left(x \right)}}{\left(1 - \tan{\left(x \right)}\right)^{2}}$$$ 关于$$$x$$$的积分

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

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

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

设$$$u=1 - \tan{\left(x \right)}$$$。

则$$$du=\left(1 - \tan{\left(x \right)}\right)^{\prime }dx = - \sec^{2}{\left(x \right)} dx$$$ (步骤见»)，并有$$$\sec^{2}{\left(x \right)} dx = - du$$$。

因此，

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

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

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

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

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

回忆一下 $$$u=1 - \tan{\left(x \right)}$$$:

$$s {\color{red}{u}}^{-1} = s {\color{red}{\left(1 - \tan{\left(x \right)}\right)}}^{-1}$$

因此，

$$\int{\frac{s \sec^{2}{\left(x \right)}}{\left(1 - \tan{\left(x \right)}\right)^{2}} d x} = \frac{s}{1 - \tan{\left(x \right)}}$$

化简：

$$\int{\frac{s \sec^{2}{\left(x \right)}}{\left(1 - \tan{\left(x \right)}\right)^{2}} d x} = - \frac{s}{\tan{\left(x \right)} - 1}$$

加上积分常数：

$$\int{\frac{s \sec^{2}{\left(x \right)}}{\left(1 - \tan{\left(x \right)}\right)^{2}} d x} = - \frac{s}{\tan{\left(x \right)} - 1}+C$$

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