Answer
$$\frac{1}{{4\sqrt 3 }}\ln \left| {\frac{{2{e^x} + \sqrt 3 }}{{2{e^x} - \sqrt 3 }}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{e^x}}}{{3 - 4{e^{2x}}}}dx} \cr
& {\text{write the integrand as}} \cr
& = \int {\frac{{{e^x}}}{{3 - {{\left( {2{e^x}} \right)}^2}}}dx} \cr
& {\text{Make an appropiate }}u{\text{ - substitution }} \cr
& u = 2{e^x},\,\,\,\,\,\,\,du = 2{e^x}dx,\,\,\,\,\,\,dx = \frac{{du}}{{2{e^x}}} \cr
& {\text{write in terms of }}u \cr
& \int {\frac{{{e^x}}}{{3 - {{\left( {2{e^x}} \right)}^2}}}dx} = \int {\frac{{{e^x}}}{{3 - {u^2}}}\left( {\frac{{du}}{{2{e^x}}}} \right)} \cr
& = \frac{1}{2}\int {\frac{1}{{3 - {u^2}}}du} \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral}} \cr
& {\text{By formula 69}} \cr
& \left( {69} \right):\,\,\,\,\,\,\int {\frac{{du}}{{{u^2} - {a^2}}} = \frac{1}{{2a}}\ln \left| {\frac{{u + a}}{{u - a}}} \right| + C} \cr
& {\text{take }}a = \sqrt 3 ,\,\, \cr
& \frac{1}{2}\int {\frac{1}{{3 - {u^2}}}du} = \frac{1}{{4\left( {\sqrt 3 } \right)}}\ln \left| {\frac{{u + \sqrt 3 }}{{u - \sqrt 3 }}} \right| + C \cr
& {\text{simplifying}} \cr
& = \frac{1}{{4\sqrt 3 }}\ln \left| {\frac{{u + \sqrt 3 }}{{u - \sqrt 3 }}} \right| + C \cr
& {\text{write in terms of }}x{\text{; replace }}2{e^x}{\text{ for }}u \cr
& = \frac{1}{{4\sqrt 3 }}\ln \left| {\frac{{2{e^x} + \sqrt 3 }}{{2{e^x} - \sqrt 3 }}} \right| + C \cr} $$
