Solution
= Solution
Regularity at zero gives
$$
r^2\phi'(r)=\int_0^r\frac{s^2}{1+s^2},ds
=r-\arctan r.
$$
Integrating once more,
$$
\phi(r)=\frac12\log(1+r^2)+\frac{\arctan r}{r}+C,
$$
where $\arctan r/r$ has removable <limit> one at zero. At $r=1$, the nonconstant terms sum to $\tfrac12\log2+\tfrac\pi4$. Hence
$$
\boxed{
\phi(r)=\frac12\log(1+r^2)+\frac{\arctan r}{r}
-1-\frac12\log2-\frac\pi4}.
$$