# Evaluate this integral

1. Oct 10, 2013

### utkarshakash

1. The problem statement, all variables and given/known data
$\displaystyle \int_0^{\infty} \dfrac{xlnx}{(1+x^2)^2} dx$

2. Relevant equations

3. 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.

2. Oct 10, 2013

### SteamKing

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

3. Oct 11, 2013

### utkarshakash

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.

4. Oct 12, 2013

### Ray Vickson

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.

5. Oct 12, 2013

### utkarshakash

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?
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