$\int{ {e}^{x} } \mathrm{d} x+\int{ \frac{ 1 }{ {e}^{x} } } \mathrm{d} x$
Use $\int{ {e}^{x} } \mathrm{d} x={e}^{x}$ to evaluate the integral${e}^{x}+\int{ \frac{ 1 }{ {e}^{x} } } \mathrm{d} x$
Evaluate the indefinite integral${e}^{x}-\frac{ 1 }{ {e}^{x} }$
Add the constant of integration $C \in ℝ$$\begin{array} { l }{e}^{x}-\frac{ 1 }{ {e}^{x} }+C,& C \in ℝ\end{array}$