For an integral say $$\int \tanh x dx= \log \cosh x \tag1$$ is it possible to add a constant say one specifically to the r.h.s logarithm of (1) giving $\log (\cosh x+1)$ with a systematic way of changing the integral on the l.h.s logarithm of (1)?
If not why?
We can obtain $\log(\cosh(x)+1)$ on the RHS by changing the integrand function since $\displaystyle\frac{\mathrm{d}}{\mathrm{d}x}\log(\cosh(x)+1)=\frac{\sinh(x)}{\cosh(x)+1}=\frac{\cosh(x)-1}{\sinh{x}}=\tanh\left(\frac{x}{2}\right)$.
This follows from representing the hyperbolic functions as exponentials:
$\displaystyle \begin{aligned}\frac{\sinh(x)}{\cosh(x)+1} &=\dfrac{\frac{e^{x}-e^{-x}}{2}}{\frac{e^{x}+e^{-x}}{2}+1} \\ &=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}+2} \\ &=\frac{(e^{\frac{x}{2}}-e^{-\frac{x}{2}})(e^{\frac{x}{2}}+e^{-\frac{x}{2}})}{(e^{\frac{x}{2}}+e^{\frac{x}{2}})(e^{\frac{x}{2}}+e^{\frac{x}{2}})} \\ &=\frac{e^{\frac{x}{2}}-e^{-\frac{x}{2}}}{e^{\frac{x}{2}}+e^{\frac{x}{2}}} \\ &=\tanh\left(\frac{x}{2}\right). \end{aligned}$
(Alternatively see here)
