The Calculation of an improper integral

Convergence:

First, let $f(x)=\frac{x \ln x}{(x^2+1)^2}$ then $f$ is decrease on $x>t$ for some $t\in\mathbb{R}$

Use $$x>\ln x \Rightarrow f(x)\leq\frac{x^2}{(x^2+1)^2}$$

and we know

$$(x^2+1)^2=x^4+2x^2+1\geq x^4 \Rightarrow \frac{1}{(x^2+1)^2}\leq\frac{1}{x^4}$$

Apply this for our first inequality, then we get $$0\leq f(x)\leq\frac{1}{x^2}\quad (x\geq1)$$

and we know that improper integral converge.

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Computing: \begin{align} &\int_0^\infty\frac{x\ln x}{(x^2+1)^2}dx=\int_0^1\frac{x\ln x} {(x^2+1)^2}dx+\int_1^\infty\frac{x\ln x}{(x^2+1)^2}dx \\ &=\int_0^1\frac{x\ln x}{(x^2+1)^2}dx-\int_0^1\frac{\ln t}{(\frac{1}{t^2}+1)^2t^3}dt \\ &=\int_0^1\frac{x\ln x}{(x^2+1)^2}dx-\int_0^1\frac{t\ln t}{(\frac{1}{t^2}+1)^2t^4}dt \\ &=\int_0^1\frac{x\ln x}{(x^2+1)^2}dx-\int_0^1\frac{x\ln x}{(x+1)^2}dx \\ &=0 \end{align}

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Substituted $x=\frac{1}{t}$ the second integral one at first line.

Yes, the improper integral is convergent, but your computation is not correct. Note that $$\begin{align} \int \frac{x \ln(x)}{(x^2+1)^2} dx& = -\frac{\ln(x)}{2(x^2+1)}+ \int\frac{1}{2x(x^2+1)}dx\\ &= -\frac{\ln(x)}{2(x^2+1)}+\frac{1}{2}\int\left(\frac{1}{x}-\frac{x}{x^2+1}\right)dx\\ &=-\frac{\ln(x)}{2(x^2+1)}+\frac{\ln(x)}{2}-\frac{\ln(1+x^2)}{4}+c\\ &=\frac{x^2\ln(x)}{2(x^2+1)}-\frac{\ln(1+x^2)}{4}+c. \end{align}$$ which can be extended by continuity in $[0,+\infty)$. Therefore $$\int_0^{+\infty} \frac{x \ln(x)}{(x^2+1)^2} dx= \left[\frac{x^2\ln(x)}{2(x^2+1)}-\frac{\ln(1+x^2)}{4}\right]_0^{+\infty}=0-0=0.$$

P.S. As regards the limit as $x\to +\infty$, note that $$\begin{align}\frac{x^2\ln(x)}{2(x^2+1)}-\frac{\ln(1+x^2)}{4}&=-\frac{\ln(x)}{2(x^2+1)} +\frac{ \ln(x^2) - \ln (x^2+1)}{4} \\&=-\frac{\ln(x)}{2(x^2+1)}+\frac{1}{4}\ln\left(\frac{x^2}{x^2+1}\right)\to 0+\ln(1)=0.\end{align}$$