Answer
$$\frac{1}{2}\ln \left( {{e^{2x}} + {e^{ - 2x}}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{e^{2x}} - {e^{ - 2x}}}}{{{e^{2x}} + {e^{ - 2x}}}}} dx \cr
& {\text{Integrate using the substitution method}} \cr
& u = {e^{2x}} + {e^{ - 2x}},{\text{ }}du = \left( {2{e^{2x}} - 2{e^{ - 2x}}} \right)dx,{\text{ }}dx = 2\left( {{e^{2x}} - {e^{ - 2x}}} \right)du \cr
& \frac{1}{2}dx = \left( {{e^{2x}} - {e^{ - 2x}}} \right)du \cr
& {\text{Substitute}} \cr
& \int {\frac{{{e^{2x}} - {e^{ - 2x}}}}{{{e^{2x}} + {e^{ - 2x}}}}} dx = \int {\frac{1}{u}\left( {\frac{1}{2}} \right)} du \cr
& = \frac{1}{2}\int {\frac{1}{u}du} \cr
& {\text{Integrate}} \cr
& = \frac{1}{2}\ln \left| u \right| + C \cr
& {\text{Write in terms of }}x \cr
& = \frac{1}{2}\ln \left| {{e^{2x}} + {e^{ - 2x}}} \right| + C \cr
& {e^{2x}} + {e^{ - 2x}} > 0,{\text{ then}} \cr
& = \frac{1}{2}\ln \left( {{e^{2x}} + {e^{ - 2x}}} \right) + C \cr} $$
