The Calculation of an improper integral For the integral $$\int_{0}^{\infty} \frac{x \ln x}{(x^2+1)^2} dx $$
I want to verify from the convergence then to calculate the integral!


*

*For the convergence, simply we can say $$ \frac{x \ln x}{(x^2+1)^2} \sim \frac{1 }{x^3 \ln^{-1} x}$$
then the integral converge because $\alpha=3 > 1$. Is this true?

*To calculate the integral, using the integration by parts where $u = \ln x$ and $dv = \frac{x \ln x}{(x^2+1)^2} dx$.
So, $$\int_{0}^{\infty} \frac{x \ln x}{(x^2+1)^2} dx =  \frac{- \ln x}{2(x^2+1)}- \frac{1}{4x^2} +\frac{\ln |x|}{2} ~\Big|_{0}^{\infty}  $$
and this undefined while it should converge to $0$ ! what I missed? 
I found an error in the calculation so the integral = So, $$\int_{0}^{\infty} \frac{x \ln x}{(x^2+1)^2} dx =  \frac{- \ln x}{2(x^2+1)}+ \frac{1}{8} \Big( \ln x^2 - \ln (x^2+1) \Big) ~\Big|_{0}^{\infty}  $$
and it's still undefined!

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.

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}

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