Integral of $$$\frac{1}{\sqrt{x^{2} - 4}}$$$

Find $$$\int \frac{1}{\sqrt{x^{2} - 4}}\, dx$$$.

Let $$$x=2 \cosh{\left(u \right)}$$$.

Then $$$dx=\left(2 \cosh{\left(u \right)}\right)^{\prime }du = 2 \sinh{\left(u \right)} du$$$ (steps can be seen »).

Also, it follows that $$$u=\operatorname{acosh}{\left(\frac{x}{2} \right)}$$$.

Integrand becomes

$$$\frac{1}{\sqrt{x^{2} - 4}} = \frac{1}{\sqrt{4 \cosh^{2}{\left( u \right)} - 4}}$$$

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

$$$\frac{1}{\sqrt{4 \cosh^{2}{\left( u \right)} - 4}}=\frac{1}{2 \sqrt{\cosh^{2}{\left( u \right)} - 1}}=\frac{1}{2 \sqrt{\sinh^{2}{\left( u \right)}}}$$$

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

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

Thus,

$${\color{red}{\int{\frac{1}{\sqrt{x^{2} - 4}} d x}}} = {\color{red}{\int{1 d u}}}$$

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

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

Recall that $$$u=\operatorname{acosh}{\left(\frac{x}{2} \right)}$$$:

$${\color{red}{u}} = {\color{red}{\operatorname{acosh}{\left(\frac{x}{2} \right)}}}$$

Therefore,

$$\int{\frac{1}{\sqrt{x^{2} - 4}} d x} = \operatorname{acosh}{\left(\frac{x}{2} \right)}$$

Add the constant of integration:

$$\int{\frac{1}{\sqrt{x^{2} - 4}} d x} = \operatorname{acosh}{\left(\frac{x}{2} \right)}+C$$

$$$\int \frac{1}{\sqrt{x^{2} - 4}}\, dx = \operatorname{acosh}{\left(\frac{x}{2} \right)} + C$$$A