Answer
$$\frac{{{e^{2t}}}}{{2\sqrt {{e^{4t}} + 1} }} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{e^{2t}}}}{{{{\left( {1 + {e^{4t}}} \right)}^{3/2}}}}} dt \cr
& {\text{Let }}u = {e^{2t}},\,\,\,du = 2{e^{2t}}dt \cr
& {\text{Apply the substitution}} \cr
& \int {\frac{{{e^{2t}}}}{{{{\left( {1 + {e^{4t}}} \right)}^{3/2}}}}} dt = \frac{1}{2}\int {\frac{{du}}{{{{\left( {1 + {u^2}} \right)}^{3/2}}}}} \cr
& {\text{The integrand contains the form }}{a^2} + {x^2} \cr
& 1 + {u^2} \to a = 1 \cr
& {\text{Use the change of variable }}u = a\tan \theta \cr
& u = \tan \theta ,\,\,\,du = {\sec ^2}\theta d\theta \cr
& \frac{1}{2}\int {\frac{{du}}{{{{\left( {1 + {u^2}} \right)}^{3/2}}}}} = \frac{1}{2}\int {\frac{{{{\sec }^2}\theta }}{{{{\left( {1 + {{\tan }^2}\theta } \right)}^{3/2}}}}d\theta } \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}\int {\frac{{{{\sec }^2}\theta }}{{{{\left( {{{\sec }^2}\theta } \right)}^{3/2}}}}d\theta } \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}\int {\frac{1}{{\sec \theta }}d\theta } \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}\int {\cos \theta d\theta } \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}\sin \theta + C \cr
& {\text{Write in terms of }}u \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}\left( {\frac{u}{{\sqrt {{u^2} + 1} }}} \right) + C \cr
& {\text{Where }}u = {e^{2t}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{e^{2t}}}}{{2\sqrt {{e^{4t}} + 1} }} + C \cr} $$
