Answer
$$y=x\arctan \frac{x}{2}-\ln(x^2+4)+c$$
Work Step by Step
Since $$
y^{\prime}=\arctan \frac{x}{2}
$$
Then
\begin{align*}
\frac{dy}{dx}&=\arctan \frac{x}{2}\\
y&=\int\arctan \frac{x}{2} dx
\end{align*}
Integrate by parts, let
\begin{align*}
u&=\arctan \frac{x}{2}\ \ \ \ \ \ \ dv=dx\\
u&=\frac{2dx}{ x^2+ 4}\ \ \ \ \ \ \ \ \ \ v=x
\end{align*}
Then
\begin{align*}
\int\arctan \frac{x}{2} dx&=uv-\int vdu\\
&=x\arctan \frac{x}{2}-\int \frac{2xdx}{ x^2+ 4} dx\\
&=x\arctan \frac{x}{2}-\ln(x^2+4)+c
\end{align*}
Hence $$y=x\arctan \frac{x}{2}-\ln(x^2+4)+c$$
