Answer
$$2\left( {4{{\ln }^2}4 - {{\ln }^2}2} \right)=30(\ln{2})^2$$
Work Step by Step
$$\eqalign{
& \int_2^4 {\left( {1 + \ln t} \right)t\ln t} dt \cr
& {\text{Use substitution:}}\cr
&{\text{Let }}u = t\ln t,{\text{ so that }}du = \left( {1 + \ln t} \right)dt \cr
& {\text{The new limits on }}u{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}t = 4,{\text{ then }}u = 4\ln 4 \cr
& \,\,\,\,\,\,{\text{If }}t = 2,{\text{ then }}u = 2\ln 2 \cr
& {\text{Then}} \cr
& \int_2^4 {\left( {1 + \ln t} \right)t\ln t} dt = \int_{2\ln 2}^{4\ln 4} u du \cr
& {\text{integrate}} \cr
& = \left( {\frac{{{u^2}}}{2}} \right)_{2\ln 2}^{4\ln 4} \cr
& {\text{use the fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 281}}} \right) \cr
& = \frac{{{{\left( {4\ln 4} \right)}^2}}}{2} - \frac{{{{\left( {2\ln 2} \right)}^2}}}{2} \cr
& = \frac{1}{2}\left( {16{{\ln }^2}4 - 4{{\ln }^2}2} \right) \cr
& = \frac{4}{2}\left( {4{{\ln }^2}4 - {{\ln }^2}2} \right) \cr
& = 2\left( {4{{\ln }^2}4 - {{\ln }^2}2} \right) \cr} $$
