Integral-how to solve?

**kaspersky0** · October 25th 2012, 12:13 PM

Hi guys,

Can anyone help me solve this integral:

**hollywood** · October 25th 2012, 08:21 PM

I think you might make some progress by substituting u=ln(x), so du=(dx/x).

- Hollywood

**kaspersky0** · October 26th 2012, 02:47 AM

I know for that kind of progress. But what is the next step? Thnx anyway

**hollywood** · October 26th 2012, 07:16 AM

Well, this is interesting. After substituting:

$\int\frac{du}{u^3(\ln{u})^2}$

And then substituting again:

$v=lnu$

$u=e^v$
$dv=\frac{du}{u}$

we have:

$\int\frac{dv}{e^{2v}v^2}$

Now integrating by parts gives:

$\int{e^{-2v}}\frac{dv}{v^2}=\int{e^{-2v}}d\left(-\frac{1}{v}\right)$
$=-\frac{e^{-2v}}{v}-\int\left(-\frac{1}{v}\right)d(e^{-2v})$
$=-\frac{e^{-2v}}{v}-2\int\frac{e^{-2v}dv}{v}$
$=-\frac{e^{-2v}}{v}-2\int\frac{e^wdw}{w}$

and we are stuck since $\int\frac{e^wdw}{w}$ can not be integrated in terms of elementary functions. See Exponential integral - Wikipedia, the free encyclopedia

Can anyone else help? Maybe I made a mistake somewhere.

- Hollywood

**kaspersky0** · October 26th 2012, 08:13 AM

Yes, I got the same result. So I wanted to ask if anyone knows the solution...

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