I recently found a weird way to integrate the function \(tanh(x)\) only using derivatives.
Take the first derivative of \(\phi = tanh(x)\) with respect to \(x\):
\[s = \frac{d\phi}{dx} = 1-\phi^2\]Now differentiate \(s\) with respect to some dummy variable \(t\):
\[\dot{s} = -2\phi\dot{\phi}= -2 \phi s \dot{x}\]The last equality is just the chain rule. Now divide both sides by \(s\), we get
\[\frac{\dot{s}}{s} = -2 \phi \dot{x}\]At this point, you have to notice that the left side of this equation is something called the logarithmic derivative of \(s\):
\[\frac{d}{dt}ln(s) = \frac{\dot{s}}{s}\]The logarithm is well defined because \(s>0\). You also have to notice that the term \(-2\phi\dot{x}\) can be written as the derivative of an integral:
\[-2\phi\dot{x}= -2 \frac{d}{dt}\int_{0}^{x}{\phi(x)dx}\]Putting both these observations together, we get:
\[\frac{d}{dt}ln(s) = -2 \frac{d}{dt}\int_{0}^{x}{\phi(x)dx}\]Integrating with respect to both sides gives us:
\[ln(s) = -2\int_{0}^{x}{\phi(x)dx}\]Dividing through by \(-2\) we get.
\[ln(\frac{1}{\sqrt{s}})= ln(\frac{1}{sech(x)}) = ln(cosh(x))= \int_{0}^{x}{\phi(x)dx}\]So we get the final answer of:
\[\int_{0}^{x}{\phi(x)dx} =\int_{0}^{x}{tanh(x)dx}= ln(cosh(x))\]Which according to the internet is correct.
Test