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!

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
$$\begin{align}\frac{x^2\ln(x)}{2(x^2+1)}-\frac{\ln(1+x^2)}{4}&=-\frac{\ln(x)}{2(x^2+1)}
+\frac{ \ln(x^2) - \ln (x^2+1)}{4}  \\&=-\frac{\ln(x)}{2(x^2+1)}+\frac{1}{4}\ln\left(\frac{x^2}{x^2+1}\right)\to 0+\ln(1)=0.\end{align}$$
