$$$\operatorname{sech}{\left(x \right)}$$$ 的积分

求$$$\int \operatorname{sech}{\left(x \right)}\, dx$$$。

将双曲正割函数用指数$$$\operatorname{sech}\left(x\right)=\frac{2}{e^{\left(x\right)}+e^{-\left(x\right)}}$$$表示:

$${\color{red}{\int{\operatorname{sech}{\left(x \right)} d x}}} = {\color{red}{\int{\frac{2}{e^{x} + e^{- x}} d x}}}$$

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

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

Simplify:

$$2 {\color{red}{\int{\frac{1}{e^{x} + e^{- x}} d x}}} = 2 {\color{red}{\int{\frac{e^{x}}{e^{2 x} + 1} d x}}}$$

设$$$u=e^{x}$$$。

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

因此，

$$2 {\color{red}{\int{\frac{e^{x}}{e^{2 x} + 1} d x}}} = 2 {\color{red}{\int{\frac{1}{u^{2} + 1} d u}}}$$

$$$\frac{1}{u^{2} + 1}$$$ 的积分为 $$$\int{\frac{1}{u^{2} + 1} d u} = \operatorname{atan}{\left(u \right)}$$$:

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

回忆一下 $$$u=e^{x}$$$:

$$2 \operatorname{atan}{\left({\color{red}{u}} \right)} = 2 \operatorname{atan}{\left({\color{red}{e^{x}}} \right)}$$

因此，

$$\int{\operatorname{sech}{\left(x \right)} d x} = 2 \operatorname{atan}{\left(e^{x} \right)}$$

加上积分常数：

$$\int{\operatorname{sech}{\left(x \right)} d x} = 2 \operatorname{atan}{\left(e^{x} \right)}+C$$

$$$\int \operatorname{sech}{\left(x \right)}\, dx = 2 \operatorname{atan}{\left(e^{x} \right)} + C$$$A