What is the integral of (e^{-3x})/(1-e^{-3x)} ?

The integral of (e^{-3x})/(1-e^{-3x)} is 1/3 ln|1-e^{-3x}|+C

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integral of (e^{-3x})/(1-e^{-3x)}

\int \frac{e ^{- 3 x}}{1 - e ^{- 3 x}} d x

\frac{1}{3} ln 1 - e^{- 3 x} + C

Solution steps

\int \frac{e ^{- 3 x}}{1 - e ^{- 3 x}} d x

Apply u-substitution

: \int \frac{1}{3 u} d u

= \int \frac{1}{3 u} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} ln ∣ u ∣

Substitute back

u = 1 - e^{- 3 x}

= \frac{1}{3} ln 1 - e^{- 3 x}

Add a constant to the solution

= \frac{1}{3} ln 1 - e^{- 3 x} + C

9

x \to \infty lim (\frac{3}{0})

y^{^{'}} - x y = 1

x \to 9 lim (x - 1)

x^{2} - 2, 6 - x^{2}

What is the integral of (e^{-3x})/(1-e^{-3x)} ?
The integral of (e^{-3x})/(1-e^{-3x)} is 1/3 ln|1-e^{-3x}|+C