Evaluate:
$$\int \frac{1}{e^x+e^{-x}}dx$$

Let $I=\int \frac{1}{e^x+e^{-x}}dx$.Multiply numerator and denominator by $e^x$:
$$I=\int \frac{e^x}{(e^x{)}^2+1}dx$$
Let $u=e^x$, then $du=e^{x}dx$.Substituting:
$$I=\int \frac{1}{u^2+1}du$$
We know that $\int \frac{1}{1+x^2}dx={\tan }^{-1}x+c$.
$$I={\tan }^{-1}u+c$$
$$I={\tan }^{-1}(e^x)+c$$