What is the integral of 1/(zln(z)) ?

The integral of 1/(zln(z)) is ln|ln(z)|+C

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integral of 1/(zln(z))

\int \frac{1}{z ln ( z )} d z

ln ∣ ln (z) ∣ + C

Solution steps

\int \frac{1}{z ln ( z )} d z

Apply u-substitution

= \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = ln (z)

= ln ∣ ln (z) ∣

Add a constant to the solution

= ln ∣ ln (z) ∣ + C