Efficiently Evaluating \frac{1}{x\ln x} Using Integration by Parts

[courtrigrad]

How do you evaluate \frac{1}{x\ln x} by integration by parts. I tried doing this doing the following:

u = \frac{1}{\ln x}, du = \frac{-1}{x(\ln x)^{2}}, dv = \frac{1}{x}, v = \ln x. So I get:

\int udv = uv-\int vdu = 1 + \int \frac{1}{x\ln x} = \int \frac{1}{x\ln x}. I know the answer is \ln(\ln x)

Thanks

 

[semc]

hmm...its not so complicated dude...the ans is ln(lnx). Ur f(x)=(lnx)^-1 and ur f'(x) is x^-1 . get it?

 

[courtrigrad]

ok i got it