How Do You Integrate 1/[xlog(x)]?

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This is doing my head in!

I split it to 1/x * 1/log(x) and got the intergral = 1 + the intergral when using intergration by parts. :cry:

I know the answer is log[log(x)] but have no idea how you get log of a log.

Got a feeling the answer is going to really obvious!
 
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Substitute u=log(x)
 
log(x) = u, u=exp(x), x du = dx.
intergral becomes exp[-u] 1/u exp du
= 1/u du
= log u
= log[log(x)]

Cheers! :biggrin:
 
\int{frac{1}{x\ln{x}}}\d x=\int{frac{1}{\ln{x}}}\d \ln{x}
 
\int{frac{1}{x\ln{x}}}\d x=\int{frac{1}{\ln{x}}}\d \ln{x}[\tex]
 
\int{frac{1}{x\ln{x}}}\d x=\int{frac{1}{\ln{x}}}\d \ln{x}
 
Do you mean?

\int\frac{1}{x\ln{x}}}dx=\int\frac{1}{\ln{x}}d\ln{x}

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I see. Sorry for making a big mess here.
 
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