Answer
$$\frac{1}{{2e - 1}}\left( {\frac{1}{2}\ln \left( {2e} \right){{\left( {2e + 1} \right)}^2} - {e^2} - 2e - \frac{1}{2}\ln \left| {2e} \right|} \right) - \frac{5}{{8e - 4}}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \left( {x + 1} \right)\ln x{\text{ over }}\left[ {1,2e} \right] \cr
& {\text{The average value of the function is given by}} \cr
& {f_{avg}} = \frac{1}{{b - a}}\int_a^b {f\left( x \right)} dx \cr
& {f_{avg}} = \frac{1}{{2e - 1}}\int_1^{2e} {\left( {x + 1} \right)\ln x} dx \cr
& {\text{Integrate }}\int {\left( {x + 1} \right)\ln xdx} {\text{ by parts}} \cr
& {\text{Let }}u = \ln x,{\text{ }}du = \frac{1}{x}dx \cr
& dv = \left( {x + 1} \right)dx,{\text{ }}v = \frac{1}{2}{\left( {x + 1} \right)^2} \cr
& \int u dv = uv - \int v du \cr
& \int {\left( {x + 1} \right)\ln xdx} = \frac{1}{2}\ln x{\left( {x + 1} \right)^2} - \int {\frac{1}{2}{{\left( {x + 1} \right)}^2}\left( {\frac{1}{x}} \right)dx} \cr
& \int {\left( {x + 1} \right)\ln xdx} = \frac{1}{2}\ln x{\left( {x + 1} \right)^2} - \frac{1}{2}\int {\left( {\frac{{{x^2} + 2x + 1}}{x}} \right)dx} \cr
& \int {\left( {x + 1} \right)\ln xdx} = \frac{1}{2}\ln x{\left( {x + 1} \right)^2} - \frac{1}{2}\int {\left( {x + 2 + \frac{1}{x}} \right)dx} \cr
& \int {\left( {x + 1} \right)\ln xdx} = \frac{1}{2}\ln x{\left( {x + 1} \right)^2} - \frac{1}{2}\left( {\frac{{{x^2}}}{2} + 2x + \ln \left| x \right|} \right) + C \cr
& \int {\left( {x + 1} \right)\ln xdx} = \frac{1}{2}\ln x{\left( {x + 1} \right)^2} - \frac{{{x^2}}}{4} - x - \frac{1}{2}\ln \left| x \right| + C \cr
& {\text{Therefore,}} \cr
& {f_{avg}} = \frac{1}{{2e - 1}}\int_1^{2e} {\left( {x + 1} \right)\ln x} dx \cr
& {f_{avg}} = \frac{1}{{2e - 1}}\left[ {\frac{1}{2}\ln x{{\left( {x + 1} \right)}^2} - \frac{{{x^2}}}{4} - x - \frac{1}{2}\ln \left| x \right|} \right]_1^{2e} \cr
& = \frac{1}{{2e - 1}}\left[ {\frac{1}{2}\ln \left( {2e} \right){{\left( {2e + 1} \right)}^2} - \frac{{{{\left( {2e} \right)}^2}}}{4} - 2e - \frac{1}{2}\ln \left| {2e} \right|} \right] \cr
& - \frac{1}{{2e - 1}}\left[ {\frac{1}{2}\ln \left( 1 \right){{\left( {1 + 1} \right)}^2} - \frac{{{{\left( 1 \right)}^2}}}{4} - 1 - \frac{1}{2}\ln \left| 1 \right|} \right] \cr
& {\text{Simplifying}} \cr
& = \frac{1}{{2e - 1}}\left( {\frac{1}{2}\ln \left( {2e} \right){{\left( {2e + 1} \right)}^2} - {e^2} - 2e - \frac{1}{2}\ln \left| {2e} \right|} \right) - \frac{5}{{8e - 4}} \cr} $$
