Integral of $ \int \frac {\tanh(x) dx}{\tanh(x)+\operatorname{sech}(x) }$ My attempt at solution:

Also I got a question, is there a way this can be solved without using hyperbolic tangential half angle substitution? Because I don't get how you can treat hyperbolic functions as if they were trigonometric and deducing sinx and cosx from a right triangle

The integral can also be seen as the following:
\begin{align}
\int \frac{\tanh(x) \, dx}{\tanh(x) + sech(x)} &= \int \frac{\sinh(x) \, dx}{\sinh(x) + 1} = \int \left( 1 - \frac{1}{1 + \sinh(x)} \right) \, dx \\
&= x + \sqrt{2} \, \tanh^{-1}\left( \frac{1}{\sqrt{2}} \, \left(1 - \tanh\left(\frac{x}{2}\right) \right) \right)   
\end{align}
