= Logarithmic quadratic parameter integral
{title2=$\int_0^\infty\frac{a\ln(a^2+x^2)}{1+x^2}dx=\pi a\ln(1+|a|)$}
For every real parameter $a$,
$$
\boxed{F(a)=\int_0^\infty\frac{a\ln(a^2+x^2)}{1+x^2}\,dx=\pi a\ln(1+|a|),}
$$
where $F(0)=0$. For $a>0$, <differentiation under the integral sign> and a <partial fraction decomposition> give
$$
F'(a)-F(a)/a=2a^2\int_0^\infty\frac{dx}{(x^2+a^2)(1+x^2)}=\frac{\pi a}{1+a}.
$$
Solving this <first-order linear differential equation> yields $F(a)=a[\pi\ln(1+a)+C]$. To fix the constant, <dominated convergence> gives $\lim_{a\to0^+}F(a)/a=\int_0^\infty\ln(x^2)/(1+x^2)dx=0$; the last absolutely convergent <integral> changes sign under $x=1/t$. Thus $C=0$. Oddness in $a$ supplies the result for negative parameters. The initial value $F(0)=0$ alone does not determine $C$, since every solution $a[\pi\ln(1+a)+C]$ tends to zero as $a\to0^+$.
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