What is the integral of (e^x)/((e^x-1)^2) ?

The integral of (e^x)/((e^x-1)^2) is -1/(e^x-1)+C

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integral of (e^x)/((e^x-1)^2)

\int \frac{e ^{x}}{( e ^{x} - 1 ) ^{2}} d x

- \frac{1}{e ^{x} - 1} + C

Solution steps

\int \frac{e ^{x}}{( e ^{x} - 1 ) ^{2}} d x

Apply u-substitution

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

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

Apply u-substitution

: \int \frac{1}{v ^{2}} d v

= \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= - \frac{1}{v}

Substitute back

= - \frac{1}{e ^{x} - 1}

Add a constant to the solution

= - \frac{1}{e ^{x} - 1} + C

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What is the integral of (e^x)/((e^x-1)^2) ?
The integral of (e^x)/((e^x-1)^2) is -1/(e^x-1)+C