Answer
$$A = \frac{\pi }{3}$$
Work Step by Step
$$\eqalign{
& y = \frac{{4{e^x}}}{{1 + {e^{2x}}}} \cr
& {\text{The area of the region is given by}} \cr
& A = \int_0^{\ln \sqrt 3 } {\frac{{4{e^x}}}{{1 + {e^{2x}}}}} dx \cr
& {\text{Let }}u = {e^x},{\text{ }}du = {e^x}dx \cr
& {\text{The new limits of integration are}} \cr
& x = \ln \sqrt 3 ,{\text{ }}u = \sqrt 3 \cr
& x = 0,{\text{ }}u = 1 \cr
& {\text{Substitute}} \cr
& A = 4\int_0^{\ln \sqrt 3 } {\frac{{{e^x}}}{{1 + {e^{2x}}}}} dx = 4\int_1^{\sqrt 3 } {\frac{{du}}{{1 + {u^2}}}} \cr
& {\text{Integrating}} \cr
& A = 4\left[ {{{\tan }^{ - 1}}u} \right]_1^{\sqrt 3 } \cr
& A = 4\left[ {{{\tan }^{ - 1}}\sqrt 3 - {{\tan }^{ - 1}}1} \right] \cr
& A = 4\left[ {\frac{\pi }{3} - \frac{\pi }{4}} \right] \cr
& {\text{Simplify}} \cr
& A = \frac{\pi }{3} \cr} $$
