What is the integral of (e^y)/((e^y-2)ln(e^y-2)) ?

The integral of (e^y)/((e^y-2)ln(e^y-2)) is ln|ln(e^y-2)|+C

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integral of (e^y)/((e^y-2)ln(e^y-2))

\int \frac{e ^{y}}{( e ^{y} - 2 ) ln ( e ^{y} - 2 )} d y

ln ∣ ln (e^{y} - 2) ∣ + C

Solution steps

\int \frac{e ^{y}}{( e ^{y} - 2 ) ln ( e ^{y} - 2 )} d y

Apply u-substitution

: \int \frac{1}{ln ( u - 2 ) ( u - 2 )} d u

= \int \frac{1}{ln ( u - 2 ) ( u - 2 )} d u

Apply u-substitution

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

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{w} d w = ln (∣ w ∣)

= ln ∣ w ∣

Substitute back

= ln ∣ ln (e^{y} - 2) ∣

Add a constant to the solution

= ln ∣ ln (e^{y} - 2) ∣ + C

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What is the integral of (e^y)/((e^y-2)ln(e^y-2)) ?
The integral of (e^y)/((e^y-2)ln(e^y-2)) is ln|ln(e^y-2)|+C