Intergrating 1/[xlog(x)]

[euclid3.14]

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!

[Zurtex]

Substitute u=log(x)

[euclid3.14]

log(x) = u, u=exp(x), x du = dx.
intergral becomes exp[-u] 1/u exp[u] du
= 1/u du
= log u
= log[log(x)]

Cheers! :biggrin:

[sinkdeep]

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

[sinkdeep]

[tex]\int{frac{1}{x\ln{x}}}\d x=\int{frac{1}{\ln{x}}}\d \ln{x}[\tex]

[sinkdeep]

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

[shmoe]

Do you mean?

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

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[sinkdeep]

I see. Sorry for making a big mess here.