What is the integral of 1/(3z(1+ln(z))) ?

The integral of 1/(3z(1+ln(z))) is 1/3 ln|1+ln(z)|+C

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integral of 1/(3z(1+ln(z)))

\int \frac{1}{3 z ( 1 + ln ( z ) )} d z

\frac{1}{3} ln ∣ 1 + ln (z) ∣ + C

Solution steps

\int \frac{1}{3 z ( 1 + ln ( z ) )} d z

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int \frac{1}{z ( 1 + ln ( z ) )} d z

Apply u-substitution

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

= \frac{1}{3} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} ln ∣ u ∣

Substitute back

u = 1 + ln (z)

= \frac{1}{3} ln ∣ 1 + ln (z) ∣

Add a constant to the solution

= \frac{1}{3} ln ∣ 1 + ln (z) ∣ + C

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What is the integral of 1/(3z(1+ln(z))) ?
The integral of 1/(3z(1+ln(z))) is 1/3 ln|1+ln(z)|+C