Evaluation of $ \int_{0}^{\infty}\frac{x\ln x}{(1+x^2)^2}dx$ $\bf{My\; Try:}$ Let $$\displaystyle I = \int_{0}^{\infty}\frac{x\ln x}{(1+x^2)^2}\,dx = \underbrace{\int_{0}^{1}\frac{x\ln x}{(1+x^2)^2}\,dx}_{I_{1}}+\underbrace{\int_{1}^{\infty}\frac{x\ln x}{(1+x^2)^2}\,dx}_{I_{2}}.$$
Now Here $$\displaystyle I_{2} = \int_{1}^{\infty}\frac{x\ln x}{(1+x^2)^2}\,dx.$$ Put $\displaystyle x=\frac{1}{t}.$ Then $\displaystyle dx = -\frac{1}{t^2}dt$ and changing limit
We get $$\displaystyle I_{2} = \int_{1}^{0}\frac{-t^3\cdot \ln t}{(1+t^2)^2}\cdot -\frac{1}{t^2}\,dt = \int_{1}^{0}\frac{t\ln t}{(1+t^2)^2}\,dt = -\int_{0}^{1}\frac{t\ln t}{(1+t^2)^2}\,dt$$
So we get $$\displaystyle I_{2} = -\int_{0}^{1}\frac{x\ln x}{(1+x^2)^2}dt = -I_{1}\Rightarrow I_{1}+I_{2} = 0$$
So we get $$\displaystyle I = \int_{0}^{\infty}\frac{x\ln x}{(1+x^2)^2}dx = I_{1}+I_{2} =0$$
My question is, can we solve it using another method? If so then please explain.

Your solution is simple and straight to the point. Here is a longer way:
You could take the indefinite integral then apply limits to find the definite integral.
Let $v=\log x$, $u'=\frac{x}{(1+x^2)^2}$
So $v'=\frac{1}{x}$, $u=-\frac{1}{2(1+x^2)}$
$$\int\frac{{x\log x}}{(1+x^2)^2}dx=-\frac{\log x}{2(1+x^2)}+\int\frac{1}{2x(1+x^2)}dx$$
$$=-\frac{\log x}{2(1+x^2)}+\int\frac{1}{2x}-\frac{x}{2(1+x^2)}dx$$
$$=-\frac{\log x}{2(1+x^2)}+\frac{\log x}{2}-\frac{\log(1+x^2)}{4}$$
$$=-\frac{x^2\log x}{2(1+x^2)}-\frac{\log(1+x^2)}{4}$$
