Answer
$$A = 2\ln \left( {\sqrt 3 + 1} \right) + \frac{1}{2} + \sqrt 3 - \ln 5$$
Work Step by Step
$$\eqalign{
& x = 0,\,\,\,\,\,x = 4,\,\,\,\,\,y = \frac{1}{{x + 1}},\,\,\,\,\,\,y = \frac{{x - 1}}{2} \cr
& {\text{Find the points of intersection of the graphs}} \cr
& \frac{1}{{x + 1}} = \frac{{x - 1}}{2} \cr
& 2 = {x^2} - 1 \cr
& {x^2} = 3 \cr
& x = \pm \sqrt 3 \cr
& {\text{Then with the given points and the critical value }}x = \sqrt 3 {\text{ we have the intervals}} \cr
& \left( {0,\sqrt 3 } \right){\text{ and }}\left( {\sqrt 3 ,4} \right) \cr
& {\text{For the interval }}\left( {0,\sqrt 3 } \right){\text{ we have that }}\frac{1}{{x + 1}} \geqslant \frac{{x - 1}}{2}{\text{ and for the interval }}\left( {\sqrt 3 ,4} \right) \cr
& \frac{{x - 1}}{2} \geqslant \frac{1}{{x + 1}} \cr
& {\text{The area is given by}}: \cr
& A = \int_0^{\sqrt 3 } {\left( {\frac{1}{{x + 1}} - \frac{{x - 1}}{2}} \right)} dx + \int_{\sqrt 3 }^4 {\left( {\frac{{x - 1}}{2} - \frac{1}{{x + 1}}} \right)} dx \cr
& A = \int_0^{\sqrt 3 } {\left( {\frac{1}{{x + 1}} - \frac{1}{2}x + \frac{1}{2}} \right)} dx + \int_{\sqrt 3 }^4 {\left( {\frac{1}{2}x - \frac{1}{2} - \frac{1}{{x + 1}}} \right)} dx \cr
& {\text{Integrating}} \cr
& A = \left( {\ln \left| {x + 1} \right| - \frac{1}{4}{x^2} + \frac{1}{2}x} \right)_0^{\sqrt 3 } + \left( {\frac{1}{4}{x^2} - \frac{1}{2}x - \ln \left| {x + 1} \right|} \right)_{\sqrt 3 }^4 \cr
& A = \left( {\ln \left| {\sqrt 3 + 1} \right| - \frac{1}{4}{{\left( {\sqrt 3 } \right)}^2} + \frac{1}{2}\left( {\sqrt 3 } \right)} \right) - \left( {\ln \left| {0 + 1} \right| - \frac{1}{4}{{\left( 0 \right)}^2} + \frac{1}{2}\left( 0 \right)} \right) \cr
& + \left( {\frac{1}{4}{{\left( 4 \right)}^2} - \frac{1}{2}\left( 4 \right) - \ln \left| {4 + 1} \right|} \right) - \left( {\frac{1}{4}{{\left( {\sqrt 3 } \right)}^2} - \frac{1}{2}\left( {\sqrt 3 } \right) - \ln \left| {\sqrt 3 + 1} \right|} \right) \cr
& \cr
& A = \ln \left| {\sqrt 3 + 1} \right| - \frac{3}{4} + \frac{{\sqrt 3 }}{2} + \left( {4 - 2 - \ln 5} \right) - \left( {\frac{3}{4} - \frac{{\sqrt 3 }}{2} - \ln \left| {\sqrt 3 + 1} \right|} \right) \cr
& A = \ln \left| {\sqrt 3 + 1} \right| - \frac{3}{4} + \frac{{\sqrt 3 }}{2} + 4 - 2 - \ln 5 - \frac{3}{4} + \frac{{\sqrt 3 }}{2} + \ln \left| {\sqrt 3 + 1} \right| \cr
& A = 2\ln \left( {\sqrt 3 + 1} \right) + \frac{1}{2} + \sqrt 3 - \ln 5 \cr} $$
