$\int_0^\infty\frac{\sqrt{x}}{(x+1)^2}\,dx$

Use the keyhole contour about the positive real axis for \[ F(z)=\frac{z^{1/2}}{(1+z)^2}, \] with the branch \(0<\arg z<2\pi\). On the upper bank, \(z^{1/2}=\sqrt x\); on the lower bank, \(z^{1/2}=-\sqrt x\), and the reversed orientation makes the two contributions add. The circular arcs vanish, so the contour integral tends to \[ 2I, \qquad I=\int_0^\infty\frac{\sqrt x}{(1+x)^2}\,dx. \]

The only pole inside the contour is the double pole at \(z=-1\). Since on our branch \((-1)^{1/2}=i\), \[ \operatorname{Res}_{z=-1}F =\left.\frac{d}{dz}z^{1/2}\right|_{z=-1} =\frac1{2i}=-\frac i2. \] Hence the residue theorem gives \[ 2I=2\pi i\left(-\frac i2\right)=\pi, \] and therefore \[ \boxed{I=\frac\pi2}. \]