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
 A: 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.
A: Firstly, note that the idea does not fully work. 
$
\def\lfrac#1#2{{\large\frac{#1}{#2}}}
\def\Int{{\displaystyle\int}}
$
After you differentiate you lose some information about the original function. Even if the derivative is continuous, integrating it will give you a class of functions that differ by a constant. So you still need to check at least one point to determine the constant, and you still need to verify that the derivative is continuous, because there are strange functions, like $\Big( \mathbb{R}\ x \mapsto \cases{ x^2 \cos(\lfrac1{x^2}) & if $x \ne 0$ \\ 0 & otherwise } \Big)$, which is differentiable on $\mathbb{R}$ but whose derivative cannot be integrated on any interval containing $0$.
With that in mind, by symmetry it suffices to prove that $\Int \lfrac1{x\sqrt{x^2+1}}\ dx = -{\ln}\Big( \lfrac1x + \lfrac{\sqrt{1+x^2}}{x} \Big) + C$ for some constant $C$, where $x > 0$.
Note the identity $\lfrac1{x\sqrt{1+x^2}} = \lfrac1x - \lfrac{x}{1+x^2+\sqrt{1+x^2}} = \lfrac1x - \lfrac{x \div \sqrt{1+x^2}}{1+\sqrt{1+x^2}}$ for any real $x > 0$.
Thus $\Int \lfrac1{x\sqrt{x^2+1}}\ dx = \ln(x) - \ln(1+\sqrt{1+x^2}) + C$ for some constant $C$.
A: we know,
$$\sinh^{-1}\left(\dfrac{1}{x}\right)=\ csch^{-1}x=\ln\left(\dfrac{1}{x}+\sqrt{\dfrac{1}{x^2}+1}\right),x\neq0$$
now, let 
$\sinh^{-1}\left(\dfrac{1}{x}\right)=y$
$\sinh y=\dfrac{1}{x}$ 
differentiate w.r.t $x$ on both sides
$\cosh  y \ \dfrac{dy}{dx}=\dfrac{-1}{x^2}$
$\dfrac{dy}{dx}=-\dfrac{1}{x^2 \ cosh \left(\sinh^{-1}\left(\dfrac{1}{x}\right)\right)}=-\dfrac{1}{x^2 \sqrt{\dfrac{1}{x^2}+1}}=-\dfrac{1}{x\sqrt{{x^2}+1}}$
from above we can infer 
$\displaystyle\int\dfrac{-1}{x\sqrt{{x^2}+1}}dx=y=\sinh^{-1}\left(\dfrac{1}{x}\right)=\ csch^{-1}x=\ln\left(\dfrac{1}{x}+\sqrt{\dfrac{1}{x^2}+1}\right)$
