What is the integral of 1/(u(ln(u)-1)) ?

The integral of 1/(u(ln(u)-1)) is ln|ln(u)-1|+C

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integral of 1/(u(ln(u)-1))

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

ln ∣ ln (u) - 1 ∣ + C

Solution steps

\int \frac{1}{u ( ln ( u ) - 1 )} 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

v = ln (u) - 1

= ln ∣ ln (u) - 1 ∣

Add a constant to the solution

= ln ∣ ln (u) - 1 ∣ + C

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What is the integral of 1/(u(ln(u)-1)) ?
The integral of 1/(u(ln(u)-1)) is ln|ln(u)-1|+C