Answer
$r = \frac{{{e^t}}}{{\ln t}} + \frac{C}{{\ln t}}$
Work Step by Step
$$\eqalign{
& t\ln t\frac{{dr}}{{dt}} + r = t{e^t} \cr
& {\text{Divide both sides by }}t\ln t \cr
& \frac{{dr}}{{dt}} + \frac{1}{{t\ln t}}r = \frac{{t{e^t}}}{{t\ln t}} \cr
& \frac{{dr}}{{dt}} + \frac{1}{{t\ln t}}r = \frac{{{e^t}}}{{\ln t}} \cr
& {\text{The differential equation is in the form }}\frac{{dr}}{{dt}} + P\left( t \right)r = Q\left( t \right) \cr
& {\text{With }}P\left( t \right) = \frac{1}{{t\ln t}}{\text{ and }}Q\left( t \right) = \frac{{{e^t}}}{{\ln t}} \cr
& {\text{Find the integrating factor }}I\left( t \right) = {e^{\int {P\left( t \right)} dt}} \cr
& I\left( t \right) = {e^{\int {\frac{1}{{t\ln t}}dt} }} = {e^{\ln \left| {\ln t} \right|}} = \ln t \cr
& {\text{Multiply the differential equation by the integrating factor}} \cr
& \left( {\frac{{dr}}{{dt}} + \frac{1}{{t\ln t}}r} \right)\ln t = \left( {\frac{{{e^t}}}{{\ln t}}} \right)\ln t \cr
& \ln t\frac{{dr}}{{dt}} + \frac{1}{t}r = {e^t} \cr
& {\text{Write the left side in the form }}\frac{d}{{dt}}\left[ {I\left( t \right)r} \right] \cr
& \frac{d}{{dt}}\left[ {r\ln t} \right] = {e^t} \cr
& d\left[ {r\ln t} \right] = {e^t}dt \cr
& {\text{Integrate both sides}} \cr
& r\ln t = {e^t} + C \cr
& {\text{Solve for }}r \cr
& r = \frac{{{e^t}}}{{\ln t}} + \frac{C}{{\ln t}} \cr} $$
