Answer
$$ r=\frac{e^t}{\ln t} +\frac{C}{\ln t}$$
Work Step by Step
Given $$t \ln t \frac{d r}{d t}+r=t e^{t}$$
Rewrite the equation
$$ \frac{d r}{d t}+\frac{1}{t\ln t}r=\frac{t e^{t} }{t\ln t}$$
Since the equation is linear and $p(t)=\frac{1}{t\ln t} $ and $q(t)=\frac{t e^{t} }{t\ln t}$, then
\begin{align*}
\mu &=e^{\int p(t)}dt\\
&= e^{\int (\frac{1}{t\ln t})dt}\\
&=e^{ \ln \ln t}\\
&=\ln t
\end{align*}
Hence the general solution given by
\begin{align*}
\mu(t)r&=\int \mu(t)q(t)dt\\
(\ln t) r &= \int \ln t \frac{t e^{t} }{t\ln t} dt\ \ \\
&=\int e^t dt\ \ \\
&=e^t +C
\end{align*}
Then
$$ r=\frac{e^t}{\ln t} +\frac{C}{\ln t}$$
