$$$\frac{1}{\sinh^{2}{\left(x \right)} \cosh^{2}{\left(x \right)}}$$$ 的积分

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

将分子和分母同时乘以 $$$\frac{1}{\cosh^{2}{\left(x \right)}}$$$，并将 $$$\frac{\cosh^{2}{\left(x \right)}}{\sinh^{2}{\left(x \right)}}$$$ 转换为 $$$\frac{1}{\tanh^{2}{\left(x \right)}}$$$:

$${\color{red}{\int{\frac{1}{\sinh^{2}{\left(x \right)} \cosh^{2}{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{1}{\cosh^{4}{\left(x \right)} \tanh^{2}{\left(x \right)}} d x}}}$$

取出两个双曲余弦项，并使用公式 $$$\cosh^{2}{\left(x \right)}=\frac{1}{1 - \tanh^{2}{\left(x \right)}}$$$ 将其他双曲余弦用双曲正切表示:

$${\color{red}{\int{\frac{1}{\cosh^{4}{\left(x \right)} \tanh^{2}{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{1 - \tanh^{2}{\left(x \right)}}{\cosh^{2}{\left(x \right)} \tanh^{2}{\left(x \right)}} d x}}}$$

设$$$u=\tanh{\left(x \right)}$$$。

则$$$du=\left(\tanh{\left(x \right)}\right)^{\prime }dx = \operatorname{sech}^{2}{\left(x \right)} dx$$$ (步骤见»)，并有$$$\operatorname{sech}^{2}{\left(x \right)} dx = du$$$。

因此，

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

Expand the expression:

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

逐项积分：

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

应用常数法则 $$$\int c\, du = c u$$$，使用 $$$c=1$$$：

$$\int{\frac{1}{u^{2}} d u} - {\color{red}{\int{1 d u}}} = \int{\frac{1}{u^{2}} d u} - {\color{red}{u}}$$

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

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

回忆一下 $$$u=\tanh{\left(x \right)}$$$:

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

因此，

$$\int{\frac{1}{\sinh^{2}{\left(x \right)} \cosh^{2}{\left(x \right)}} d x} = - \tanh{\left(x \right)} - \frac{1}{\tanh{\left(x \right)}}$$

加上积分常数：

$$\int{\frac{1}{\sinh^{2}{\left(x \right)} \cosh^{2}{\left(x \right)}} d x} = - \tanh{\left(x \right)} - \frac{1}{\tanh{\left(x \right)}}+C$$

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