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

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

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

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

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

Solution steps

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

Factor

1 - e^{- 3 x} : - (- 1 + e^{- 3 x})

= \int \frac{1}{- ( - 1 + e ^{- 3 x} )} d x

Take the constant out:

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

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

Apply u-substitution

: \int - \frac{1}{3 u ( u + 1 )} d u

= - \int - \frac{1}{3 u ( u + 1 )} 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 ( u + 1 )} d u)

Take the partial fraction of

\frac{1}{u ( u + 1 )} : \frac{1}{u} - \frac{1}{u + 1}

= - (- \frac{1}{3} \cdot \int \frac{1}{u} - \frac{1}{u + 1} d u)

Apply the Sum Rule:

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

= - (- \frac{1}{3} (\int \frac{1}{u} d u - \int \frac{1}{u + 1} d u))

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{1}{u + 1} d u = ln ∣ u + 1 ∣

= - (- \frac{1}{3} (ln ∣ u ∣ - ln ∣ u + 1 ∣))

Substitute back

u = - 1 + e^{- 3 x}

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

Simplify

- (- \frac{1}{3} (ln - 1 + e^{- 3 x} - ln - 1 + e^{- 3 x} + 1)) : \frac{1}{3} (ln - 1 + e^{- 3 x} + 3 x)

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

Add a constant to the solution

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

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