Answer
$$y=x\ln x-x+c$$
Work Step by Step
Since $$
y^{\prime}=\ln x
$$
Then
\begin{align*}
\frac{dy}{dx}&=\ln x\\
y&=\int\ln x dx
\end{align*}
Integrate by parts, let
\begin{align*}
u&=\ln x\ \ \ \ \ \ \ dv=dx\\
u&=\frac{dx}{ x}\ \ \ \ \ \ \ \ \ \ v=x
\end{align*}
Then
\begin{align*}
\int\ln x dx&=uv-\int vdu\\
&=x\ln x-\int dx\\
&=x\ln x-x+c
\end{align*}
Hence $$y=x\ln x-x+c$$
