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

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

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integral of 1/((y+1)ln(y+1))

\int \frac{1}{( y + 1 ) ln ( y + 1 )} d y

ln ∣ ln (y + 1) ∣ + C

Solution steps

\int \frac{1}{( y + 1 ) ln ( y + 1 )} d y

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 (y + 1) ∣

Add a constant to the solution

= ln ∣ ln (y + 1) ∣ + C

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