Answer
$$\frac{1}{{18}}\left[ {\frac{4}{{4 - 3{e^{2x}}}} + \ln \left| {4 - 3{e^{2x}}} \right|} \right] + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{e^{4x}}}}{{{{\left( {4 - 3{e^{2x}}} \right)}^2}}}dx} ,\,\,\,\,u = {e^{2x}} \cr
& {\text{Using the given substitution}} \cr
& u = {e^{2x}},\,\,\,\,\,\,\,du = 2{e^{2x}}dx,\,\,\,\,\,\,dx = \frac{{du}}{{2{e^{2x}}}} \cr
& {\text{write in terms of }}u \cr
& \int {\frac{{{e^{4x}}}}{{{{\left( {4 - 3{e^{2x}}} \right)}^2}}}dx} = \int {\frac{{{e^{4x}}}}{{{{\left( {4 - 3u} \right)}^2}}}\left( {\frac{{du}}{{2{e^{2x}}}}} \right)} \cr
& = \int {\frac{{{e^{2x}}}}{{{{\left( {4 - 3u} \right)}^2}}}\left( {\frac{{du}}{2}} \right)} \cr
& = \frac{1}{2}\int {\frac{u}{{{{\left( {4 - 3u} \right)}^2}}}du} \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral}} \cr
& {\text{By formula 62}} \cr
& \int {\frac{{udu}}{{{{\left( {a + bu} \right)}^2}}} = \frac{1}{{{b^2}}}\left[ {\frac{a}{{a + bu}} + \ln \left| {a + bu} \right|} \right] + C} \cr
& {\text{take }}a = 4,\,\,\,b = - 3{\text{ }} \cr
& \frac{1}{2}\int {\frac{u}{{{{\left( {4 - 3u} \right)}^2}}}du} = \frac{1}{{2{{\left( { - 3} \right)}^2}}}\left[ {\frac{4}{{4 - 3u}} + \ln \left| {4 - 3u} \right|} \right] + C \cr
& {\text{simplifying}} \cr
& \frac{1}{2}\int {\frac{u}{{{{\left( {4 - 3u} \right)}^2}}}du} = \frac{1}{{18}}\left[ {\frac{4}{{4 - 3u}} + \ln \left| {4 - 3u} \right|} \right] + C \cr
& {\text{write in terms of }}x{\text{ replace }}{e^{2x}}{\text{ for }}u \cr
& = \frac{1}{{18}}\left[ {\frac{4}{{4 - 3{e^{2x}}}} + \ln \left| {4 - 3{e^{2x}}} \right|} \right] + C \cr} $$
