Answer
$$\frac{1}{2}{\tan ^{ - 1}}\left( {{e^{2x}}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{{e^{2x}} + {e^{ - 2x}}}}} dx \cr
& {\text{Multiply the numerator and denominator by }}{e^{2x}} \cr
& = \int {\frac{{{e^{2x}}}}{{{e^{4x}} + {e^0}}}} dx \cr
& = \int {\frac{{{e^{2x}}}}{{{e^{4x}} + 1}}} dx \cr
& {\text{Let }}u = {e^{2x}},{\text{ }}du = 2{e^{2x}}dx,{\text{ }}dx = \frac{{du}}{{2{e^{2x}}}} \cr
& {\text{Substitute}} \cr
& = \int {\frac{{{e^{2x}}}}{{{u^2} + 1}}} \frac{{du}}{{2{e^{2x}}}} \cr
& = \frac{1}{2}\int {\frac{1}{{{u^2} + 1}}} du \cr
& {\text{Integrate}} \cr
& {\text{ = }}\frac{1}{2}{\tan ^{ - 1}}u + C \cr
& {\text{ = }}\frac{1}{2}{\tan ^{ - 1}}\left( {{e^{2x}}} \right) + C \cr} $$
