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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the partial fraction of

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{1}{u} d u = ln ∣ u ∣

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

\int \frac{1}{u - 1} d u = ln ∣ u - 1 ∣

= - ln ∣ u ∣ - (- \frac{1}{u}) + ln ∣ u - 1 ∣

Substitute back

u = e^{x}

= - ln ∣ e^{x} ∣ - (- \frac{1}{e ^{x}}) + ln ∣ e^{x} - 1 ∣

Simplify

- ln ∣ e^{x} ∣ - (- \frac{1}{e ^{x}}) + ln ∣ e^{x} - 1 ∣ : - x + \frac{1}{e ^{x}} + ln ∣ e^{x} - 1 ∣

= - x + \frac{1}{e ^{x}} + ln ∣ e^{x} - 1 ∣

Add a constant to the solution

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

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