Evaluating $\int ^\infty _{0} \frac{x\ln x}{(1+x^2)^2} \,dx$

\begin{align} I&=\int_0^\infty\frac{x\ln x}{(1+x^2)^2}\ dx\overset{\large x\ \mapsto\frac1x}{=}-I\\ 2I&=0\\ I&=0 \end{align}

Set $\dfrac1x=y$

$$\int ^\infty _{0} \frac{x\ln x}{(1+x^2)^2} \,dx=I=\int_\infty^0\dfrac{y^4\ln(y^{-1})}{y(1+y^2)^2}\left(\dfrac{dy}{-y^2}\right)=-\int_0^\infty\dfrac{y\ln y}{(1+y^2)^2}\ dy =-I$$

Another way:

Integrate by parts

$$\int\ln x\cdot\dfrac x{(1+x^2)^2}\ dx=\ln x\int\dfrac x{(1+x^2)^2}\ dx-\int\left(\dfrac{d(\ln x)}{dx}\int\dfrac x{(1+x^2)^2}\ dx\right)dx$$

$$=-\dfrac{\ln x}{2(1+x^2)}+\int\dfrac{dx}{2x(1+x^2)}$$

Again $$\int\dfrac{dx}{x(1+x^2)}=\int\dfrac{(x^2+1-x^2)\ dx}{x(1+x^2)}=\int\dfrac{dx}x-\int\dfrac{x\ dx}{1+x^2}=\dfrac12\ln\dfrac{x^2}{1+x^2}$$