Answer
$$\ln \left( {\frac{{21}}{{13}}} \right)$$
Work Step by Step
$$\eqalign{
& \int_{ - \ln 3}^{\ln 3} {\frac{{{e^x}}}{{{e^x} + 4}}} dx \cr
& u = {e^x} + 4,\,\,\,du = {e^x}dx \cr
& {\text{for }}x = \ln 3,\,\,\,\,\,\,\,\,\,\,u = {e^{\ln 3}} + 4 = 7 \cr
& {\text{for }}x = - \ln 3,\,\,\,\,\,\,u = {e^{ - \ln 3}} + 4 = \frac{{13}}{3} \cr
& {\text{Thus}}{\text{,}} \cr
& \int_{ - \ln 3}^{\ln 3} {\frac{{{e^x}}}{{{e^x} + 4}}} dx = \int_{13/3}^7 {\frac{1}{u}} du\, \cr
& {\text{integrating}} \cr
& = \left[ {\ln \left| u \right|} \right]_{13/3}^7 \cr
& {\text{evaluating the limits}} \cr
& = \ln \left| 7 \right| - \ln \left| {\frac{{13}}{3}} \right| \cr
& = \ln \left( {\frac{7}{{3/3}}} \right) \cr
& = \ln \left( {\frac{{21}}{{13}}} \right) \cr} $$
