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

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

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

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

- 1 + e^{- x} + ln 1 - e^{- x} + C

Solution steps

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

Apply u-substitution

: \int - \frac{u}{1 - u} d u

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Apply u-substitution

: \int - \frac{- v + 1}{v} d v

= - \int - \frac{- v + 1}{v} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - (- \int \frac{- v + 1}{v} d v)

Expand

\frac{- v + 1}{v} : - 1 + \frac{1}{v}

= - (- \int - 1 + \frac{1}{v} d v)

Apply the Sum Rule:

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

= - (- (- \int 1 d v + \int \frac{1}{v} d v))

\int 1 d v = v

\int \frac{1}{v} d v = ln ∣ v ∣

= - (- (- v + ln ∣ v ∣))

Substitute back

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

Simplify

= - 1 + e^{- x} + ln 1 - e^{- x}

Add a constant to the solution

= - 1 + e^{- x} + ln 1 - e^{- x} + C

y^{^{'}} = - \frac{4 t}{y}

x \to - 1 lim (\frac{x ^{2} - 5}{1 - x})

\frac{d}{d x} (arctan (x - x^{2} + 1))

\int (1 + sin (x))^{2} d x

(x^{2} + y^{2} + x) d x + x y d y = 0

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