= Solution
Introduce the <Mellin transform> integral $J(s)=\int_0^\infty x^{s-1}/(1+x^2)\,dx$, valid for $0<\operatorname{Re}s<2$. A <keyhole contour> for $z^{s-1}/(1+z^2)$ with $0<\arg z<2\pi$ has residues $i^{s-1}/(2i)$ and $(-i)^{s-1}/(-2i)$, with the latter argument $3\pi/2$. The two sides of the positive-axis cut give $(1-e^{2\pi i(s-1)})J(s)$. The small and large arcs vanish in this strip; simplifying the <residue theorem> identity yields
$$
J(s)=\frac\pi2\csc\frac{\pi s}{2}.
$$
Differentiation under the integral is valid on compact substrips because the extra $\log x$ is dominated at both endpoints. Therefore
$$
J'(s)=-\frac{\pi^2}{4}\csc\frac{\pi s}{2}\cot\frac{\pi s}{2}.
$$
At $s=3/2$, the requested values are
$$
\boxed{\int_0^\infty\frac{x^{1/2}\log x}{1+x^2}\,dx=\frac{\pi^2}{2\sqrt2},\qquad
\int_0^\infty\frac{x^{1/2}}{1+x^2}\,dx=\frac\pi{\sqrt2}}.
$$
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