Answer
$$\frac{1}{4}\ln \left( {4{e^x} + 6} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{e^x}}}{{4{e^x} + 6}}} dx \cr
& {\text{Let }}u = 4{e^x} + 6 \Rightarrow du = 4{e^x}dx \cr
& {\text{Apply the substitution}} \cr
& \int {\frac{{{e^x}}}{{4{e^x} + 6}}} dx = \int {\frac{{{e^x}}}{u}} \left( {\frac{{du}}{{4{e^x}}}} \right) \cr
& {\text{ = }}\frac{1}{4}\int {\frac{{du}}{u}} \cr
& {\text{Integrate}} \cr
& {\text{ = }}\frac{1}{4}\ln \left| u \right| + C \cr
& {\text{Back - substitute }}u = 4{e^x} + 6 \cr
& {\text{ = }}\frac{1}{4}\ln \left| {4{e^x} + 6} \right| + C \cr
& 4{e^x} + 6{\text{ is always posittive, then}} \cr
& {\text{ = }}\frac{1}{4}\ln \left( {4{e^x} + 6} \right) + C \cr} $$
