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

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

设$$$x=\cosh{\left(u \right)}$$$。

则$$$dx=\left(\cosh{\left(u \right)}\right)^{\prime }du = \sinh{\left(u \right)} du$$$（步骤见»）。

此外，可得$$$u=\operatorname{acosh}{\left(x \right)}$$$。

所以，

$$$\frac{1}{\left(x^{2} - 1\right)^{\frac{3}{2}}} = \frac{1}{\left(\cosh^{2}{\left( u \right)} - 1\right)^{\frac{3}{2}}}$$$

利用恒等式 $$$\cosh^{2}{\left( u \right)} - 1 = \sinh^{2}{\left( u \right)}$$$：

$$$\frac{1}{\left(\cosh^{2}{\left( u \right)} - 1\right)^{\frac{3}{2}}}=\frac{1}{\left(\sinh^{2}{\left( u \right)}\right)^{\frac{3}{2}}}$$$

假设$$$\sinh{\left( u \right)} \ge 0$$$，我们得到如下结果：

$$$\frac{1}{\left(\sinh^{2}{\left( u \right)}\right)^{\frac{3}{2}}} = \frac{1}{\sinh^{3}{\left( u \right)}}$$$

因此，

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

将被积函数用双曲余割表示:

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

$$$\operatorname{csch}^{2}{\left(u \right)}$$$ 的积分为 $$$\int{\operatorname{csch}^{2}{\left(u \right)} d u} = - \coth{\left(u \right)}$$$:

$${\color{red}{\int{\operatorname{csch}^{2}{\left(u \right)} d u}}} = {\color{red}{\left(- \coth{\left(u \right)}\right)}}$$

回忆一下 $$$u=\operatorname{acosh}{\left(x \right)}$$$:

$$- \coth{\left({\color{red}{u}} \right)} = - \coth{\left({\color{red}{\operatorname{acosh}{\left(x \right)}}} \right)}$$

因此，

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

加上积分常数：

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

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