Find $\int_{0}^{\infty} \frac{\log(x) }{\sqrt{x} (x+1)^{2}}\,dx$ Need solve the next integral
$$\int_{0}^{\infty} \frac{\log(x) }{\sqrt{x} (x+1)^{2}}\,dx$$
Tried something with Laurent’s series, but i can’t conclude anything.
Thanks

METHODOLGY $1$:  CONTOUR INTEGRATION
Enforce the substitution $x\mapsto x^2$ to find that
$$\int_0^\infty \frac{\log(x)}{\sqrt x(x+1)^2}\,dx=4\int_0^\infty \frac{\log(x)}{(x^2+1)^2}\,dx\tag1$$

Let $f(z)$ be given by 
$$f(z)=\oint_C \frac{\log^2(z)}{(z^2+1)^2}\,dz\tag2$$
where we choose to cut the plane along the positive real axis and where $C$ is the classical keyhole contour.

We then have from $(2)$
$$\begin{align}
\int_0^\infty \frac{\log^2(x)-\left(\log(x)+i2\pi\right)^2}{(x^2+1)^2}\,dx&=2\pi i \text{Res}\left(\frac{\log^2(z)}{(z^2+1)^2}, z=e^{i\pi/2}\right)\\\\&+2\pi i \text{Res}\left(\frac{\log^2(z)}{(z^2+1)^2}, z=e^{i3\pi/2}\right)\tag3\end{align}$$

The left-hand side of $(3)$ becomes
$$\begin{align}
\int_0^\infty \frac{\log^2(x)-\left(\log(x)+i2\pi\right)^2}{(x^2+1)^2}\,dx&=-i4\pi\int_0^\infty \frac{\log(x)}{(x^2+1)^2}\,dx\\\\& +4\pi^2\int_0^\infty \frac{1}{(x^2+1)^2}\,dx\tag4
\end{align}$$

Note that the imaginary part of the right-hand side of $(4)$ is $-\pi$ times the integral of interest on the right-hand side of $(1)$.  Thus, we find that 
$$\begin{align}
\int_0^\infty \frac{\log(x)}{\sqrt x(x+1)^2}\,dx&=-2\text{Re}\left(\text{Res}\left(\frac{\log^2(z)}{(z^2+1)^2}, z=e^{i\pi/2}\right)\right)\\\\
&-2 \text{Re}\left(\text{Res}\left(\frac{\log^2(z)}{(z^2+1)^2}, z=e^{i3\pi/2}\right)\right)\\\\
&=-2\left(-\frac\pi4+\frac{3\pi}{4}\right)\\\\
&=-\pi
\end{align}$$


METHODOLGY $2$:  REAL ANALYSIS ONLY
We begin by enforcing the substitution $x\mapsto \tan(x)$ in the integral on the right-hand side of $(1)$ to reveal 
$$\begin{align}
4\int_0^\infty \frac{\log(x)}{(x^2+1)^2}\,dx&=4\int_0^{\pi/2}\cos^2(x) \log(\tan(x))\,dx\\\\
&=4\int_0^{\pi/2}\cos^2(x) \log(\sin(x))\,dx\\\\&-4\int_0^{\pi/2}\cos^2(x) \log(\cos(x))\,dx\tag5 \\\\
&=4\int_0^{\pi/2}(2\cos^2(x)-1)\log(\sin(x))\,dx\tag6\\\\
&=4\int_0^{\pi/2}\cos(2x)\log(\sin(x))\,dx\tag7\\\\
&=4\left(-\int_0^{\pi/2}\cos^2(x)\,dx\right)\tag8\\\\
&=-\pi
\end{align}$$
as expected.
In going from $(5)$ to $(6)$ we made use of the transformation $x\mapsto \pi/2 -x$ in the second integral on the right-hand side of $(5)$.
In going from $(7)$ to $(8)$, we used integration by parts with $u=\log(\sin(x))$ and $v=\sin(x)\cos(x)$
