## Thomas' Calculus 13th Edition

$\tan^{-1}(e^{z})+C$
Multiplying numerator and denominator with $e^{z}$, we get $\int \frac{dz}{e^{z}+e^{-z}}=\int \frac{e^{z}}{e^{2z}+1}dz$ Now, substitute $u=e^{z}$ so that $dz=\frac{du}{e^{z}}$. Then, $\int \frac{e^{z}}{e^{2z}+1}dz=\int \frac{e^{z}}{u^{2}+1}\frac{du}{e^{z}}=\int\frac{1}{u^{2}+1}du$ $=\tan^{-1}u+C=\tan^{-1}(e^{z})+C$