What is the integral of (e^{-bx})/(1-e^{-2bx)} ?

The integral of (e^{-bx})/(1-e^{-2bx)} is -1/(2b)(ln|e^{-bx}+1|-ln|e^{-bx}-1|)+C

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integral of (e^{-bx})/(1-e^{-2bx)}

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

- \frac{1}{2 b} (ln e^{- b x} + 1 - ln e^{- b x} - 1) + C

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

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

= - \frac{1}{b} \cdot \int \frac{1}{1 - v ^{2}} d v

Use the common integral:

\int \frac{1}{1 - v ^{2}} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= - \frac{1}{b} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

Substitute back

= - \frac{1}{b} (\frac{ln e ^{- b x} + 1}{2} - \frac{ln e ^{- b x} - 1}{2})

Simplify

- \frac{1}{b} (\frac{ln e ^{- b x} + 1}{2} - \frac{ln e ^{- b x} - 1}{2}) : - \frac{1}{2 b} (ln e^{- b x} + 1 - ln e^{- b x} - 1)

= - \frac{1}{2 b} (ln e^{- b x} + 1 - ln e^{- b x} - 1)

Add a constant to the solution

= - \frac{1}{2 b} (ln e^{- b x} + 1 - ln e^{- b x} - 1) + C

\int x ln (6 x) d x

\frac{2}{x ^{2}}

\frac{d}{d x} (ln ((x^{2})^{2}))

(- π, 0), (- \frac{π}{2}, 4)

\frac{2}{y}

What is the integral of (e^{-bx})/(1-e^{-2bx)} ?
The integral of (e^{-bx})/(1-e^{-2bx)} is -1/(2b)(ln|e^{-bx}+1|-ln|e^{-bx}-1|)+C