integrate in a keyhole contour and show that $$ \int_{0}^{\infty}{\frac{x^{1/2}\log {x}}{1+x^2}dx}=\pi^2/\sqrt{8}$$ and $$ \int_{0}^{\infty}{\frac{x^{1/2}}{1+x^2}dx}=\pi/\sqrt{2}$$
We use the principal branch of $$x^{0.5} $$ and $$\log x$$ I tried solving using $${\frac{\log {x}}{1+x^2}dx}$$ but couldn't get.
$$ \int_0^{\infty} \frac{x^{s-1}}{1+x^2} \, dx = \frac{1}{2} \int_0^{\infty} \frac{x^{s/2-1}}{1+x} \, dx. $$ Now use $$ \frac{1}{1+x} = \int_0^{\infty} e^{-\alpha (1+x)} \, d\alpha $$ and interchange the order of integration to find $$ \frac{1}{2} \int_0^{\infty} \frac{x^{s/2-1}}{1+x} \, dx = \frac{1}{2} \int_0^{\infty} e^{-\alpha} \left( \int_0^{\infty} x^{s/2-1}e^{-\alpha x} \, dx \right) \, d\alpha \\ = \frac{ \Gamma(s/2)}{2} \int_0^{\infty} e^{-\alpha} \alpha^{-s/2} \, d\alpha \\ = \frac{\Gamma(s/2)}{2} \Gamma(1-s/2) = \tfrac{1}{2}\pi \csc{\tfrac{1}{2}\pi s}, $$ using the reflection formula.
You can now find your second integral by setting $s = 3/2$, and the first by differentiating with respect to $s$ and setting both $s=3/2$.