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!
 A: 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.
A: 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
$$\begin{align}\frac{x^2\ln(x)}{2(x^2+1)}-\frac{\ln(1+x^2)}{4}&=-\frac{\ln(x)}{2(x^2+1)}
+\frac{ \ln(x^2) - \ln (x^2+1)}{4}  \\&=-\frac{\ln(x)}{2(x^2+1)}+\frac{1}{4}\ln\left(\frac{x^2}{x^2+1}\right)\to 0+\ln(1)=0.\end{align}$$
A: A series expansion around $x=0$ gives
$$ \frac{x \log( x)}{(x^2+1)^2}=x \log (x)-2 x^3 \log (x)+O\left(x^5\right)$$ So, no problem at the lower bound.
For infinitely large values of $x$
$$ \frac{x \log( x)}{(x^2+1)^2}=\frac{\log \left({x}\right)}{x^3}+O\left(\frac{1}{x^5}\right)$$ and no problem either.
A: The limit of $\frac{x \log{x}}{(x^2+1)^2}$ has the same behavior as $\frac{\log{x}}{x^3}$.
You can use L'Hôpital's rule and see that it converges to $\frac{1}{3x^3}$, which goes to $0$ as $x$ approaches infinity.
Using integration by parts with $u=\log{x}$ and $\frac{dv}{dx} = \frac{x}{(x^2+1)^2}$. We know that $\frac{du}{dx} = \frac{1}{x} $, and that $ v = \int_0^x \frac{x}{(x^2+1)^2} dx$. Using the substitution $x=\tan{y}$, we obtain $ v = \int_0^{\tan^{-1}{x}} \frac{\tan{y}}{(\tan^2{y}+1)^2} dx
= \int_{{\pi}/{2}}^{\tan^{-1}{x}}\sin{y}\cdot \cos^3{y} \space dx = -\frac{1}{4}\cos^4{y} \space|_{\pi/2}^{\tan^{-1}{x}} = -\frac{1}{4(x^2+1)^2}
$.
The integral becomes $\int_0^\infty \frac{x \log{x}}{(x^2+1)^2} dx= uv -\int_0^\infty v \frac{du}{dx}dx
=-\frac{\log{x}}{4(x^2+1)^2}|_0^\infty + \int_0^\infty \frac{1}{4x(x^2+1)^2}
=\frac{1}{4}-\frac{1}{4}=0
$.
A: Concerning the calculation of the integral.
Using the substitution $x=\frac{1}{t}$ you get 
$$I = \int_0^{\infty} \frac{x \ln x}{(x^2+1)^2} dx \stackrel{x=\frac{1}{t}}{=}\int_{\infty}^0 \frac{-\frac{\ln t}{t}}{\left(\frac{1}{t^2}+1\right)^2}\left(-\frac{1}{t^2}\right)\;dt $$ $$= -\int_0^{\infty}\frac{t \ln t}{(t^2+1)^2} \; dt = -I$$
Hence, $\boxed{I = 0}$.
A: $$I=\int_0^\infty\frac{x\ln(x)}{(x^2+1)^2}dx$$
$u=\ln(x)\Rightarrow dx=xdu$
$$I=\int_{-\infty}^\infty\frac{x^2u}{(x^2+1)^2}dx=\int_{-\infty}^\infty\frac{u}{(e^{2u}+1)^2}du-\int_{-\infty}^\infty\frac u{e^{2u}+1}du$$
