Integral of $$$\frac{1}{\sqrt{x^{2} - 1}}$$$

Find $$$\int \frac{1}{\sqrt{x^{2} - 1}}\, dx$$$.

Let $$$x=\cosh{\left(u \right)}$$$.

Then $$$dx=\left(\cosh{\left(u \right)}\right)^{\prime }du = \sinh{\left(u \right)} du$$$ (steps can be seen »).

Also, it follows that $$$u=\operatorname{acosh}{\left(x \right)}$$$.

Thus,

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

Use the identity $$$\cosh^{2}{\left( u \right)} - 1 = \sinh^{2}{\left( u \right)}$$$:

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

Assuming that $$$\sinh{\left( u \right)} \ge 0$$$, we obtain the following:

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

Therefore,

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

Apply the constant rule $$$\int c\, du = c u$$$ with $$$c=1$$$:

$${\color{red}{\int{1 d u}}} = {\color{red}{u}}$$

Recall that $$$u=\operatorname{acosh}{\left(x \right)}$$$:

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

Therefore,

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

Add the constant of integration:

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

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