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