Answer
$$A = 6\ln \frac{3}{2} + 2{e^{ - 1}} + 2e - 6$$
Work Step by Step
$$\eqalign{
& x = - 1,\,\,\,\,\,x = 1,\,\,\,\,\,y = {e^x},\,\,\,\,\,\,y = 3 - {e^x} \cr
& {\text{Find the points of intersection of the graphs}} \cr
& {e^x} = 3 - {e^x} \cr
& 0 = 3 - {e^x} - {e^x} \cr
& - 2{e^x} = - 3 \cr
& x = \ln \frac{3}{2} \cr
& {\text{Then with the given points and the critical value }}x = \sqrt 3 {\text{ we have the intervals}} \cr
& \left( { - 1,\ln \frac{3}{2}} \right){\text{ and }}\left( {\ln \frac{3}{2},1} \right) \cr
& {\text{For the interval }}\left( { - 1,\ln \frac{3}{2}} \right){\text{ we have that }}3 - {e^x} > {e^x}{\text{ and }} \cr
& {\text{For the interval }}\left( {\ln \frac{3}{2},1} \right){\text{ }}{e^x} > 3 - {e^x} \cr
& \cr
& {\text{The area is given by}}: \cr
& A = \int_{ - 1}^{\ln 3/2} {\left( {3 - {e^x} - {e^x}} \right)} dx + \int_{\ln 3/2}^1 {\left( {{e^x} - 3 + {e^x}} \right)} dx \cr
& A = \int_{ - 1}^{\ln 3/2} {\left( {3 - 2{e^x}} \right)} dx + \int_{\ln 3/2}^1 {\left( {2{e^x} - 3} \right)} dx \cr
& {\text{Integrating}} \cr
& A = \left( {3x - 2{e^x}} \right)_{ - 1}^{\ln 3/2} + \left( {2{e^x} - 3x} \right)_{\ln 3/2}^1 \cr
& A = \left( {3\left( {\ln \frac{3}{2}} \right) - 2{e^{\ln 3/2}}} \right) - \left( {3\left( { - 1} \right) - 2{e^{ - 1}}} \right) + \left( {2{e^1} - 3\left( 1 \right)} \right) - \left( {2{e^{\ln 3/2}} - 3\left( {\ln 3/2} \right)} \right) \cr
& A = \left( {3\ln \frac{3}{2} - 3} \right) - \left( { - 3 - 2{e^{ - 1}}} \right) + \left( {2e - 3} \right) - \left( {3 - 3\ln \frac{3}{2}} \right) \cr
& A = 3\ln \frac{3}{2} - 3 + 3 + 2{e^{ - 1}} + 2e - 3 - 3 + 3\ln \frac{3}{2} \cr
& A = 6\ln \frac{3}{2} + 2{e^{ - 1}} + 2e - 6 \cr} $$