What is the integral of (e^{-0.05x})/(1-e^{-0.05x)} ?

The integral of (e^{-0.05x})/(1-e^{-0.05x)} is 20ln|1-e^{-0.05x}|+C

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integral of (e^{-0.05x})/(1-e^{-0.05x)}

\int \frac{e ^{- 0.05 x}}{1 - e ^{- 0.05 x}} d x

20 ln 1 - e^{- 0.05 x} + C

Solution steps

\int \frac{e ^{- 0.05 x}}{1 - e ^{- 0.05 x}} d x

Apply u-substitution

: \int \frac{1}{0.05 u} d u

= \int \frac{1}{0.05 u} d u

Take the constant out:

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

= \frac{1}{0.05} \cdot \int \frac{1}{u} d u

Simplify

= 20 \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 20 ln ∣ u ∣

Substitute back

u = 1 - e^{- 0.05 x}

= 20 ln 1 - e^{- 0.05 x}

Add a constant to the solution

= 20 ln 1 - e^{- 0.05 x} + C

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What is the integral of (e^{-0.05x})/(1-e^{-0.05x)} ?
The integral of (e^{-0.05x})/(1-e^{-0.05x)} is 20ln|1-e^{-0.05x}|+C