# Integral-how to solve?

• Oct 25th 2012, 12:13 PM
kaspersky0
Integral-how to solve?
Hi guys,

Can anyone help me solve this integral:

Attachment 25401
• Oct 25th 2012, 08:21 PM
hollywood
Re: Integral-how to solve?
I think you might make some progress by substituting u=ln(x), so du=(dx/x).

- Hollywood
• Oct 26th 2012, 02:47 AM
kaspersky0
Re: Integral-how to solve?
I know for that kind of progress. But what is the next step? Thnx anyway ;)
• Oct 26th 2012, 07:16 AM
hollywood
Re: Integral-how to solve?
Well, this is interesting. After substituting:

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

And then substituting again:

$\displaystyle v=lnu$

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

we have:

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

Now integrating by parts gives:

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

and we are stuck since $\displaystyle \int\frac{e^wdw}{w}$ 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
• Oct 26th 2012, 08:13 AM
kaspersky0
Re: Integral-how to solve?
Yes, I got the same result. So I wanted to ask if anyone knows the solution...