Use integration by parts $\int^{\infty}_{0} \frac{x \cdot \ln x}{(1+x^2)^2}dx$ $$I=\int^{\infty}_{0} \frac{x \cdot \ln x}{(1+x^2)^2}dx$$
Clearly $$-2I=\int^{\infty}_{0} \ln x \cdot \frac{-2x }{(1+x^2)^2} dx$$
My attempt :
$$-2I=\left[ \ln x \cdot \left(\frac{1}{1+x^2}\right)\right]^\infty_0 - \int^{\infty}_{0} \left(\frac{1}{1+x^2}\right) \cdot \frac{1}{x} dx$$
$$-2I=\left[ \ln x \cdot \left(\frac{1}{1+x^2}\right)\right]^\infty_0 - \int^{\infty}_{0} \frac{1}{x(1+x^2)}  dx$$
$$-2I=\left[ \ln x \cdot \left(\frac{1}{1+x^2}\right)\right]^\infty_0 - \int^{\infty}_{0} \frac{1+x^2-x^2}{x(1+x^2)}  dx$$
$$-2I=\left[ \ln x \cdot \left(\frac{1}{1+x^2}\right)\right]^\infty_0 - \int^{\infty}_{0} \frac{1}{x}+ \frac{1}{2}\int^{\infty}_{0} \frac{2x}{1+x^2} dx$$
$$-2I=\left[ \ln x \cdot \left(\frac{1}{1+x^2}\right)\right]^\infty_0 -\left[ \ln x -\frac{1}{2}\cdot \ln(1+x^2) \right]^\infty_0 $$
$$-2I=\left[  \frac{\ln x}{1+x^2}\right]^\infty_0 -\left[\ln \left (\frac{x}{\sqrt{1+x^2}} \right) \right]^\infty_0 $$
How can I evaluate the last limits ? 

A faster approach:
$$ J(a,b)=\int_{0}^{+\infty}\frac{x^b}{a^2+x^2}\,dx = a^{b-1} \frac{\pi}{2\cos\left(\frac{\pi b}{2}\right)}\qquad (a>0,b\in(-1,1))\tag{1}$$
is a consequence of the reflection formula for the $\Gamma$ function and Euler's beta function properties.
By considering:
$$ \lim_{b\to 1^-}\lim_{a\to 1^-}-\frac{1}{2}\frac{\partial^2 J}{\partial a \partial b} \tag{2}$$
we recover the value of our integral by differentiation under the integral sign. That gives:

$$ I=\int_{0}^{+\infty}\frac{x\log x}{(1+x^2)^2}\,dx = \lim_{b\to 1^-} -\frac{\pi}{8}\cdot \frac{2+\pi(b-1)\tan\left(\frac{\pi b}{2}\right)}{ \cos\left(\frac{\pi b}{2}\right)}=\color{red}{0}\tag{3}$$

that also follows from a symmetry argument:

$$ I = \int_{0}^{1}\frac{x\log x}{(1+x^2)^2}\,dx +\int_{0}^{1}\frac{\frac{1}{x}\log\left(\frac{1}{x}\right)}{x^2\left(1+\frac{1}{x^2}\right)^2}\,dx=\int_{0}^{1}0\,dx = \color{red}{0}.\tag{4} $$

