= Solution
First suppose $\alpha>0$. <Differentiation under the integral sign>, justified on compact positive parameter <intervals> by an integrable bound, gives
$$
f'(\alpha)=\int_0^\infty\frac{\ln(\alpha^2+x^2)}{1+x^2}\,dx+2\alpha^2\int_0^\infty\frac{dx}{(x^2+\alpha^2)(x^2+1)}.
$$
The first <integral> is $f(\alpha)/\alpha$. For $\alpha\ne1$, use the <partial fraction decomposition>
$$
\frac1{(x^2+\alpha^2)(x^2+1)}=\frac1{1-\alpha^2}\left(\frac1{x^2+\alpha^2}-\frac1{x^2+1}\right).
$$
The standard <inverse tangent> <antiderivatives> then give
$$
\int_0^\infty\frac{dx}{(x^2+\alpha^2)(x^2+1)}=\frac1{1-\alpha^2}\left(\frac\pi{2\alpha}-\frac\pi2\right)=\frac\pi{2\alpha(1+\alpha)}.
$$
The same value holds at $\alpha=1$ by <continuity>, or by direct <integration>. Hence the proposed <first-order linear differential equation> is valid on the positive parameter range:
$$
\boxed{f'(\alpha)-\frac{f(\alpha)}\alpha=\frac{\pi\alpha}{1+\alpha},\qquad\alpha>0.}
$$
Divide by $\alpha$ to recognise a <quotient rule> derivative:
$$
\left(\frac{f(\alpha)}\alpha\right)'=\frac\pi{1+\alpha},\qquad f(\alpha)=\alpha[\pi\ln(1+\alpha)+C].
$$
To fix $C$, the value $f(0)=0$ alone is insufficient, since every displayed solution tends to zero there. Instead consider
$$
\lim_{\alpha\to0^+}\frac{f(\alpha)}\alpha=\int_0^\infty\frac{\ln(x^2)}{1+x^2}\,dx.
$$
This limit follows from the <dominated convergence theorem>: for $0<\alpha\leq1$ and $0<x<1$, the logarithm's <absolute value> is bounded by $2|\ln x|+\ln2$; for $x\geq1$, it is at most $\ln(1+x^2)$. After division by $1+x^2$, these are integrable. The limiting <improper integral> is absolutely convergent, and the <substitution rule> $x=1/t$ changes it into its own negative. Its value is therefore zero, giving $C=0$. Thus
$$
\boxed{\int_0^\infty\frac{\alpha\ln(\alpha^2+x^2)}{1+x^2}\,dx=\pi\alpha\ln(1+\alpha),\qquad\alpha>0.}
$$
The question calls $\alpha$ a real parameter, but this last formula and the proposed differential equation do not hold on the entire real axis. The defining <integral> is odd in $\alpha$, because its logarithm depends on $\alpha^2$. Applying the positive-parameter result to $|\alpha|$ therefore gives the correct extension
$$
\boxed{f(\alpha)=\pi\alpha\ln(1+|\alpha|),\qquad\alpha\in\mathbb R,}
$$
with $f(0)=0$. In particular, for $\alpha<0$, the right-hand side of the differential equation must be $-\pi\alpha/(1-\alpha)$; for every $\alpha\ne0$ the unified expression is $\pi|\alpha|/(1+|\alpha|)$. The <logarithmic quadratic parameter integral> therefore provides the requested identity on its valid range and the corrected answer elsewhere.
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