Answer
$$\frac{1}{2}{e^x} + \frac{1}{2}\ln \left| {\cos {e^x} + \sin {e^x}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{e^x}}}{{1 - \tan {e^x}}}} dx \cr
& {\text{Let }}u = {e^x}.{\text{ Then }}du = {e^x}dx,{\text{ substituting you have}} \cr
& \int {\frac{{{e^x}}}{{1 - \tan {e^x}}}} dx = \int {\frac{{du}}{{1 - \tan u}}} \cr
& {\text{From the table of integrals in the back of the book}} \cr
& \int {\frac{{du}}{{1 - \tan u}}} = \frac{1}{2}\left( {u + \ln \left| {\cos u + \sin u} \right|} \right) + C \cr
& {\text{Write in terms of }}x,{\text{ }}u = {e^x} \cr
& = \frac{1}{2}\left( {{e^x} + \ln \left| {\cos {e^x} + \sin {e^x}} \right|} \right) + C \cr
& = \frac{1}{2}{e^x} + \frac{1}{2}\ln \left| {\cos {e^x} + \sin {e^x}} \right| + C \cr} $$
