Evaluating $\int\frac{\sqrt{x^2-1}}x\mathrm dx$

Question

How can one evaluate the integral $$\int\frac{\sqrt{x^2-1}}x\mathrm dx$$?

I tried substituting $x = \cosh t$ but got stuck at $$\int\frac{\sinh^2t}{\cosh t}\mathrm dt$$

Any hints?

Answer 1

Put $X^2=x^2-1$ then $x^2=X^2+1$.

$2x\mathrm dx=2X\mathrm dX$.

$$\int\frac{\sqrt{x^2-1}}{x}\mathrm dx=\int\frac{X^2}{x^2}\mathrm dX=\int\frac{X^2}{X^2+1}\mathrm dX$$

Answer 2

Let $ x = \sec u $. Then, $ \mathrm{d}x = \sec u \tan u \, \mathrm{d}u $. Then, the integral becomes $$ \int \tan^2 u \, \mathrm{d}u = \int \left( \sec^2 u - 1 \right) \, \mathrm{d}u = \tan u - u + \mathcal{C}. $$ Then, you can substitute back and finish.

Answer 3

I'll use the hyperbolic substitution you made. (Why not?) Of importance is the hyperbolic dual of the Pythagorean identity, $\cosh^2 x - \sinh^2 x = 1$. Then, one can see that: $$\frac{\sinh^2 t}{\cosh t} = \frac{\cosh^2 t - 1}{\cosh t} $$

This makes your integral: $$\int \cosh t - \operatorname{sech} t\,dt$$

If you know your hyperbolic trig integrals as well as most people know their "normal" trig integrals, you're home free.

Hint: $$\int\operatorname{sech} t\,dt = 2\arctan\left(\tanh\left(\frac{t}{2}\right)\right) + C$$ (According to Wolfram.)

Answer 4

$$ \begin{aligned}\int \frac{\sqrt{x^{2}-1}}{x} d x =& \int \frac{x^{2}-1}{x \sqrt{x^{2}-1}} d x \\ =& \int \frac{x^{2}-1}{x^{2}} d\left(\sqrt{x^{2}-1}\right) \\ =& \int\left(1-\frac{1}{x^{2}}\right) d\left(\sqrt{x^{2}-1}\right) \\ =& \sqrt{x^{2}-1}-\int \frac{d\left(\sqrt{x^{2}-1}\right)}{\left(\sqrt{x^{2}-1}\right)^{2}+1} \\ =& \sqrt{x^{2}-1}-\tan ^{-1}\left(\sqrt{x^{2}-1}\right)+C \end{aligned} $$