= Solution
The stated derivative formula needs a domain qualification. The original PDF calls $\alpha$ real, but the integral is even in $\alpha$; its derivative therefore cannot equal the printed positive expression on both sides of zero.
First take $\alpha>0$. The <improper integral> converges, with an $O(x^2)$ integrand near zero and $O(\ln x/x^2)$ decay at infinity. On compact positive parameter intervals, an integrable bound also justifies <differentiation under the integral sign>:
$$
E'(\alpha)=\int_0^\infty\frac{2\alpha x^2}{(1+\alpha^2x^2)(1+x^2)}\,dx.
$$
For $\alpha\ne1$, the <partial fraction decomposition> is
$$
\frac{x^2}{(1+\alpha^2x^2)(1+x^2)}
=\frac1{\alpha^2-1}\left(\frac1{1+x^2}-\frac1{1+\alpha^2x^2}\right).
$$
Integrating the rational functions gives
$$
E'(\alpha)=\frac{2\alpha}{\alpha^2-1}\left(\frac\pi2-\frac\pi{2\alpha}\right)
=\boxed{\frac\pi{1+\alpha}\quad(\alpha>0).}
$$
At $\alpha=1$, the same result follows by continuity, or directly from $2\int_0^\infty x^2/(1+x^2)^2\,dx=\pi/2$.
Integrating gives $E(\alpha)=\pi\ln(1+\alpha)+C$ on the positive half-line. To fix $C$, take $\alpha\to0^+$: for $0\le\alpha\le1$ the integrand is bounded by $\ln(1+x^2)/(1+x^2)$, an integrable function. The <dominated convergence theorem> gives $E(\alpha)\to E(0)=0$, so $C=0$. Evenness then extends the <logarithmic parameter integral with Cauchy weight> to its full stated domain:
$$
\boxed{E(\alpha)=\pi\ln(1+|\alpha|)\quad(\alpha\in\mathbb R).}
$$
For negative $\alpha$, the derivative is $-\pi/(1-\alpha)$. At zero the right and left <derivatives> are $\pi$ and $-\pi$, so $E$ is not <differentiable> there. The pointwise derivative of the integrand at zero is zero, but interchanging it with the infinite <integral> at that parameter is invalid. Finally, substituting $\alpha=1$ gives the requested special value
$$
\boxed{\int_0^\infty\frac{\ln(1+x^2)}{1+x^2}\,dx=\pi\ln2.}
$$
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