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

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

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integral of 1/(zln(z)ln(ln(z)))

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

ln ∣ ln (ln (z)) ∣ + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ln ( u )} d u

= \int \frac{1}{u ln ( u )} d u

Apply u-substitution

: \int \frac{1}{v} d v

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

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= ln ∣ v ∣

Substitute back

= ln ∣ ln (ln (z)) ∣

Add a constant to the solution

= ln ∣ ln (ln (z)) ∣ + C

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