Answer
$$A = \ln 5$$
Work Step by Step
$$\eqalign{
& {\text{The area is given by}} \cr
& A = \int_{ - \ln 5}^{\ln 5} {\frac{1}{{1 + {e^x}}}dx} \cr
& {\text{Multiply the numerator and denominator by }}{e^{ - x}} \cr
& A = \int_{ - \ln 5}^{\ln 5} {\frac{{{e^{ - x}}}}{{{e^{ - x}} + 1}}dx} \cr
& A = - \int_{ - \ln 5}^{\ln 5} {\frac{{ - {e^{ - x}}}}{{{e^{ - x}} + 1}}dx} \cr
& {\text{Integrating}} \cr
& A = - \left[ {\ln \left( {{e^{ - x}} + 1} \right)} \right]_{ - \ln 5}^{\ln 5} \cr
& A = - \left[ {\ln \left( {{e^{ - \ln 5}} + 1} \right)} \right] + \left[ {\ln \left( {{e^{\ln 5}} + 1} \right)} \right] \cr
& A = - \left[ {\ln \left( {\frac{1}{5} + 1} \right)} \right] + \left[ {\ln \left( {5 + 1} \right)} \right] \cr
& A = - \ln \frac{6}{5} + \ln 6 \cr
& A = \ln \frac{6}{{6/5}} \cr
& A = \ln 5 \cr} $$
