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!

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
$.
