Answer
$$\frac{{{e^{2x}} - {e^{ - 2x}}}}{2}$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{e^{2x}} - {e^{ - 2x}}}}{2}dx} \cr
& {\text{split the numerator}} \cr
& = \int {\left( {\frac{{{e^{2x}}}}{2} - \frac{{{e^{ - 2x}}}}{2}} \right)dx} \cr
& = \int {\frac{{{e^{2x}}}}{2}dx} - \int {\frac{{{e^{ - 2x}}}}{2}dx} \cr
& = \frac{1}{2}\int {{e^{2x}}dx} - \frac{1}{2}\int {{e^{ - 2x}}dx} \cr
& \operatorname{integrate} \cr
& = \frac{1}{2}\left( {\frac{1}{2}{e^{2x}}} \right) - \frac{1}{2}\left( { - \frac{1}{2}{e^{ - 2x}}} \right) + C \cr
& {\text{simplify}} \cr
& = \frac{1}{4}{e^{2x}} + \frac{1}{4}{e^{ - 2x}} + C \cr
& \cr
& {\text{check by differentiation}} \cr
& {\text{ = }}\frac{d}{{dx}}\left( {\frac{1}{4}{e^{2x}} + \frac{1}{4}{e^{ - 2x}} + C} \right) \cr
& {\text{ = }}\frac{1}{4}{e^{2x}}\left( 2 \right) + \frac{1}{4}{e^{ - 2x}}\left( 2 \right) + 0 \cr
& {\text{ = }}\frac{1}{2}{e^{2x}} - \frac{1}{2}{e^{ - 2x}} \cr
& = \frac{{{e^{2x}} - {e^{ - 2x}}}}{2} \cr} $$
