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

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

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

So,

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

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

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

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

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

So,

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

Multiply the numerator and denominator by one hyperbolic cosine and write everything else in terms of the hyperbolic sine, using the formula $$$\cosh^2\left(\alpha \right)=\sinh^2\left(\alpha \right)+1$$$ with $$$\alpha= u $$$:

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

Let $$$v=\sinh{\left(u \right)}$$$.

Then $$$dv=\left(\sinh{\left(u \right)}\right)^{\prime }du = \cosh{\left(u \right)} du$$$ (steps can be seen »), and we have that $$$\cosh{\left(u \right)} du = dv$$$.

Therefore,

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

Rewrite and split the fraction:

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

Integrate term by term:

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

Apply the constant rule $$$\int c\, dv = c v$$$ with $$$c=1$$$:

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

The integral of $$$\frac{1}{v^{2} + 1}$$$ is $$$\int{\frac{1}{v^{2} + 1} d v} = \operatorname{atan}{\left(v \right)}$$$:

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

Recall that $$$v=\sinh{\left(u \right)}$$$:

$$- \operatorname{atan}{\left({\color{red}{v}} \right)} + {\color{red}{v}} = - \operatorname{atan}{\left({\color{red}{\sinh{\left(u \right)}}} \right)} + {\color{red}{\sinh{\left(u \right)}}}$$

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

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

Therefore,

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

Add the constant of integration:

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

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