Answer
$$ - \frac{1}{4}\ln \left| {\frac{{2 + {e^{ - x}}}}{{2 - {e^{ - x}}}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{e^{ - x}}}}{{4 - {e^{ - 2x}}}}dx} \cr
& = \int {\frac{{{e^{ - x}}dx}}{{{{\left( 2 \right)}^2} - {{\left( {{e^{ - x}}} \right)}^2}}}} \cr
& {\text{substitute }}u = {e^{ - x}},{\text{ }} \cr
& du = - {e^{ - x}}dx \cr
& - du = {e^{ - x}}dx \cr
& = \int {\frac{{{e^{ - x}}dx}}{{{{\left( 2 \right)}^2} - {{\left( {{e^{ - x}}} \right)}^2}}}} = \int {\frac{{ - du}}{{{{\left( 2 \right)}^2} - {u^2}}}} \cr
& = - \int {\frac{{du}}{{{2^2} - {u^2}}}} \cr
& {\text{find the antiderivative }}\left( {{\text{formula 26 page 490}}} \right) \cr
& = - \int {\frac{{du}}{{{2^2} - {u^2}}}} = - \frac{1}{{2\left( 2 \right)}}\ln \left| {\frac{{2 + u}}{{2 - u}}} \right| + C \cr
& {\text{write in terms of }}x,{\text{ replace }}u = {e^{ - x}} \cr
& = - \frac{1}{{2\left( 2 \right)}}\ln \left| {\frac{{2 + {e^{ - x}}}}{{2 - {e^{ - x}}}}} \right| + C \cr
& = - \frac{1}{4}\ln \left| {\frac{{2 + {e^{ - x}}}}{{2 - {e^{ - x}}}}} \right| + C \cr} $$
