Chapter 8 - Techniques of Integration - 8.4 Integrals Involving Hyperbolic and Inverse Hyperbolic Functions - Exercises - Page 415: 43

$$\tanh (x/2) +C$$

Given $$\int \frac{d x}{1+\cosh x}$$ Let $u=\tanh (x/2) $ $$\cosh x=\frac{1+u^{2}}{1-u^{2}}, \quad \sinh x=\frac{2 u}{1-u^{2}}, \quad d x=\frac{2 d u}{1-u^{2}} $$ Then \begin{aligned} \int \frac{d x}{1+\cosh x} &=\int \frac{1-u^{2}}{2}\left(\frac{2 d u}{1-u^{2}}\right) \\ &=\int d u \\ &=u+C\\ &=\tanh (x/2) +C \end{aligned}