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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

= \frac{1}{b} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{b} ln ∣ u ∣

Substitute back

u = 1 - e^{- b x}

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

Add a constant to the solution

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

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

f (x) = y

\frac{- s}{s ^{2} + 25 s + 30} + \frac{1}{s}

\int e^{2 x} (1 - 3 x) d x

x \to 4 lim (e^{x})

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