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What is the antiderivative of

$$\int\frac{1}{r \ln(r)} \ dr$$

I'm trying to use substitution, but substituting $u=r$ doesn't help as that just changes the variable.

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Try using $u=\ln(r)$. – Clayton Dec 11 '12 at 5:20
One last hint: $\dfrac{1}{r \ln r} = \dfrac{1}{r} \cdot \dfrac{1}{\ln r}$. – JavaMan Dec 11 '12 at 6:07

As suggested in the comments, try the substitution $u = \ln (r)$. Then, $du = \frac{1}{r} \ dr$.

$$\int\frac{1}{r \ln(r)} \ dr = \int\frac{du}{u} = \ln(u) + C = \ln\left(\ln(r)\right) + C$$

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Consider the substitution $u=\ln(r)$,

Then $u'=\frac1r$

Have you tried that?

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