# Intergrating 1/[xlog(x)]

1. Nov 26, 2004

### 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.

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!

2. Nov 26, 2004

### Zurtex

Substitute u=log(x)

3. Nov 26, 2004

### euclid3.14

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!

4. Nov 27, 2004

### sinkdeep

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

5. Nov 27, 2004

$$\int{frac{1}{x\ln{x}}}\d x=\int{frac{1}{\ln{x}}}\d \ln{x}[\tex] 6. Nov 27, 2004 ### sinkdeep [tex]\int{frac{1}{x\ln{x}}}\d x=\int{frac{1}{\ln{x}}}\d \ln{x}$$

7. Nov 27, 2004

### shmoe

Do you mean?

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

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8. Nov 27, 2004

### sinkdeep

I see. Sorry for making a big mess here.