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 ? 

HINT:
Let $x=\tan y$
$$\implies2J=\int_0^{\pi/2}\sin2y\cdot\ln(\tan y)dy$$
Now use $$I=\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$$
$$\implies2I=\int_a^bf(x)dx+\int_a^bf(a+b-x)dx=\int_a^b\left(f(x)+f(a+b-x)\right)dx$$
