Answer
$${\text{A}} = \frac{{{e^2} + 5}}{4}$$
Work Step by Step
$$\eqalign{
& {\text{let }}y = \left( {x + 1} \right)\ln x,{\text{ the }}x{\text{ - axis from }}x = 1{\text{ to }}x = e \cr
& {\text{The area between two curves is defined by }}\int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx \cr
& {\text{where }}f{\text{ and }}g{\text{ are continuous functions and }}f\left( x \right) \geqslant g\left( x \right){\text{ on }}\left[ {a,b} \right]{\text{for this exercise }} \cr
& g\left( x \right) = 0{\text{ }}\left( {x{\text{ - axis}}} \right){\text{ and the interval }}x = 2{\text{ to }}x = 4{\text{ implies that }}a = 2{\text{ and }}b = 4 \cr
& {\text{then}} \cr
& {\text{A}} = \int_1^e {\left[ {\left( {x + 1} \right)\ln x - 0} \right]} dx \cr
& {\text{A}} = \int_1^e {\left( {x + 1} \right)\ln x} dx \cr
& {\text{setting }}\,\,\,\,\,\,u = \ln x{\text{ then }}du = \frac{1}{x}dx\,\,\,\,\,\,\,\,\,\, \cr
& {\text{and}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,dv = \left( {x + 1} \right)dx{\text{ then }}v = \frac{{{{\left( {x + 1} \right)}^2}}}{2} \cr
& {\text{Substituting these values into the formula for integration by parts}} \cr
& \int u dv = uv - \int {vdu} \cr
& \int {\left( {x + 1} \right)\ln x} dx = \frac{{{{\left( {x + 1} \right)}^2}\ln x}}{2} - \int {\frac{{{{\left( {x + 1} \right)}^2}}}{2}\left( {\frac{1}{x}} \right)} dx \cr
& {\text{integrate}} \cr
& \int {\left( {x + 1} \right)\ln x} dx = \frac{{{{\left( {x + 1} \right)}^2}\ln x}}{2} - \frac{1}{2}\int {\frac{{{x^2} + 2x + 1}}{x}} dx \cr
& \int {\left( {x + 1} \right)\ln x} dx = \frac{{{{\left( {x + 1} \right)}^2}\ln x}}{2} - \frac{1}{2}\int {\left( {x + 2 + \frac{1}{x}} \right)} dx \cr
& \int {\left( {x + 1} \right)\ln x} dx = \frac{{{{\left( {x + 1} \right)}^2}\ln x}}{2} - \frac{1}{2}\left( {\frac{{{x^2}}}{2} + 2x + \ln \left| x \right|} \right) + C \cr
& \int {\left( {x + 1} \right)\ln x} dx = \frac{{{{\left( {x + 1} \right)}^2}\ln x}}{2} - \frac{{{x^2}}}{4} - x - \frac{1}{2}\ln \left| x \right| + C \cr
& \cr
& {\text{Now find the definite integral}} \cr
& {\text{A}} = \int_1^e {\left( {x + 1} \right)\ln x} dx = \left. {\left( {\frac{{{{\left( {x + 1} \right)}^2}\ln x}}{2} - \frac{{{x^2}}}{4} - x - \frac{1}{2}\ln \left| x \right|} \right)} \right|_1^e \cr
& {\text{evaluating the limits}} \cr
& {\text{A}} = \left( {\frac{{{{\left( {e + 1} \right)}^2}\ln e}}{2} - \frac{{{e^2}}}{4} - e - \frac{1}{2}\ln \left| e \right|} \right) - \left( {\frac{{{{\left( {1 + 1} \right)}^2}\ln 1}}{2} - \frac{{{1^2}}}{4} - 1 - \frac{1}{2}\ln \left| 1 \right|} \right) \cr
& {\text{simplifying}} \cr
& {\text{A}} = \left( {\frac{{{{\left( {e + 1} \right)}^2}}}{2} - \frac{{{e^2}}}{4} - e - \frac{1}{2}} \right) - \left( {\frac{0}{2} - \frac{5}{4} - 0} \right) \cr
& {\text{A}} = \frac{{{e^2} + 2e + 1}}{2} - \frac{{{e^2}}}{4} - e - \frac{1}{2} + \frac{5}{4} \cr
& {\text{A}} = \frac{{{e^2}}}{2} + e + \frac{1}{2} - \frac{{{e^2}}}{4} - e - \frac{1}{2} + \frac{5}{4} \cr
& {\text{A}} = \frac{{{e^2}}}{4} + \frac{5}{4} \cr
& {\text{A}} = \frac{{{e^2} + 5}}{4} \cr} $$
