<s>[INST] Identity of an inverse hyperbolic function Recently I stumbled upon the identity: $$\operatorname{csch}^{-1}(x) = \ln \left( \frac 1 x + \frac{\sqrt{1+x^2}}{|x|}\right)$$
So i decided to prove this identity by taking the integral of the derivative of inverse $\operatorname{csch}(x)$. Although i tried expressing the integral : $$-\int \frac{\mathrm dx}{ |x|\sqrt{1+x^2}}$$
into different forms, I still haven't found a way to write it in terms of $$\int \frac{f'(x)}{f(x)} \, \mathrm dx$$
**I know there is a solution which involves the application of the definition of hyperbolic functions (with exponentials), but I was wandering if it was possibile with integration
 [/INST] The usual algebraic proof:  Recall that $$\sinh x = \frac{e^{x} - e^{-x}}{2}.$$  Thus, $$\operatorname{csch} x = \frac{1}{\sinh x} = \frac{2}{e^x - e^{-x}}.$$  It follows that $y = \operatorname{csch}^{-1} x$ is the solution to the equation $$x = \frac{2}{e^y - e^{-y}}.$$  To this end, we write $$e^y - e^{-y} = 2/x,$$ and multiplying through by $e^y$ gives the quadratic $$(e^y)^2 - (2/x) e^y - 1 = 0,$$ which has the solution $$e^y = \frac{(2/x) \pm \sqrt{4/x^2 + 4}}{2} = \frac{1}{x} \pm \sqrt{\frac{1}{x^2} + 1}.$$  Since $$\frac{1}{x^2} + 1 > \frac{1}{x^2},$$ if we require $y$ to be real-valued, we must choose the positive root of the quadratic; hence $$y = \operatorname{csch}^{-1} x = \log \left( \frac{1}{x} + \sqrt{\frac{1}{x^2} + 1} \right), \quad x \ne 0.$$

Consider $$I = \int \frac{dx}{x \sqrt{x^2+1}} = \int \frac{x \, dx}{x^2 \sqrt{x^2+1}}.$$  With the substitution $$u^2 = x^2 + 1, \quad 2u \, du = 2x \, dx,$$ we obtain $$\int \frac{u \, du}{u(u^2-1)} = \int \frac{du}{u^2-1} = \frac{1}{2}\left(\int \frac{du}{u-1} - \int \frac{du}{u+1}\right).$$  It follows that $$I = \frac{1}{2} \log \frac{u-1}{u+1} + C = \frac{1}{2} \log \frac{(u-1)^2}{u^2 - 1} + C = \log \frac{u-1}{x} + C = \log \frac{-1 + \sqrt{x^2+1}}{x} + C.$$  The details are left to you to complete, namely to demonstrate $C = 0$, and to refine the argument to take care of the absolute values.
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