Answer
$\tan^{-1}(e^{z})+C$
Work Step by Step
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,
$\int \frac{e^{z}}{e^{2z}+1}dz=\int \frac{e^{z}}{u^{2}+1}\frac{du}{e^{z}}=\int\frac{1}{u^{2}+1}du$
$=\tan^{-1}u+C=\tan^{-1}(e^{z})+C$
