# Evaluate this integral

Gold Member

## Homework Statement

$\displaystyle \int_0^{\infty} \dfrac{xlnx}{(1+x^2)^2} dx$

## The Attempt at a Solution

Integrating by parts and using ILATE rule

$\left[ ln \dfrac{x}{\sqrt{1+x^2}} - \dfrac{lnx}{2(1+x^2)} \right]$

Now I find the limit as x tends to infinity and get 0. But how do I evaluate limit when x tends to zero.

SteamKing
Staff Emeritus
Homework Helper
Your indefinite integration went off the rails somewhere. How did you ever get a SQRT(1+x^2) term?

Gold Member
Your indefinite integration went off the rails somewhere. How did you ever get a SQRT(1+x^2) term?

I have checked my integrals thrice and it seems that I can't find any error in it. Anyways, I'm posting here for you

$lnx \displaystyle \int \dfrac{xdx}{(1+x^2)^2} - \int \dfrac{1}{x} \int \frac{xdx}{(1+x^2)^2}\\ \frac{-lnx}{2(1+x^2)} + \frac{1}{2} \int \dfrac{dx}{x(1+x^2)} \\ \dfrac{1}{2} \left[ lnx - ln \sqrt{1+x^2} - \dfrac{lnx}{1+x^2} \right]$

If this is further simplified it turns out to be the same as posted earlier. I only missed that 1/2, but it makes no difference when limit is to be evaluated.

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

$\displaystyle \int_0^{\infty} \dfrac{xlnx}{(1+x^2)^2} dx$

## The Attempt at a Solution

Integrating by parts and using ILATE rule

$\left[ ln \dfrac{x}{\sqrt{1+x^2}} - \dfrac{lnx}{2(1+x^2)} \right]$

Now I find the limit as x tends to infinity and get 0. But how do I evaluate limit when x tends to zero.

First, check that your indefinite integral is correct, by differentiating it to see if you recover the integrand. That should always be your first step! I think your integral is incorrect; but do not take my word for it; check it for yourself.

Gold Member
First, check that your indefinite integral is correct, by differentiating it to see if you recover the integrand. That should always be your first step! I think your integral is incorrect; but do not take my word for it; check it for yourself.

OK, I made a silly mistake and now here's the correct one

$\dfrac{1}{4} \ln \dfrac{x^2}{1+x^2} - \dfrac{1}{2} \dfrac{\ln x}{1+x^2}$

I can evaluate limit when x tends to infinity but can't get it for zero. Can you please help?