= Solution
For $0<a<2$, the beta-integral evaluation gives
$$
I(a)=\int_0^\infty\frac{x^{a-1}}{1+x^2}\,dx
=\frac\pi2\csc\frac{\pi a}{2}.
$$
<Differentiation> under the <integral> sign near $a=1$ is justified by domination at zero and infinity. Therefore
$$
I^{(4)}(1)=\int_0^\infty\frac{(\log x)^4}{1+x^2}\,dx.
$$
Writing $a=1+t$ gives
$$
I(1+t)=\frac\pi2\sec\frac{\pi t}{2}.
$$
Since
$$
\sec z=1+\frac{z^2}{2}+\frac{5z^4}{24}+O(z^6),
$$
the <logarithmic moments of the Cauchy kernel> yield
$$
\boxed{
\int_0^\infty\frac{(\log x)^4}{1+x^2}\,dx
=\frac{5\pi^5}{32}}.
$$
The same expansion gives $I''(1)=\pi^3/8$, agreeing with the supplied identity.
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