What is the integral of 1/(9e^v-1) ?

The integral of 1/(9e^v-1) is ln|9e^v-1|-2ln(3)-v+C

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integral of 1/(9e^v-1)

\int \frac{1}{9 e ^{v} - 1} d v

ln ∣ 9 e^{v} - 1 ∣ - 2 ln (3) - v + C

Solution steps

\int \frac{1}{9 e ^{v} - 1} d v

Apply u-substitution

: \int \frac{1}{u ( 9 u - 1 )} d u

= \int \frac{1}{u ( 9 u - 1 )} d u

Apply u-substitution

: \int \frac{1}{w ( w + 1 )} d w

= \int \frac{1}{w ( w + 1 )} d w

Take the partial fraction of

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

= \int \frac{1}{w} - \frac{1}{w + 1} d w

Apply the Sum Rule:

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

= \int \frac{1}{w} d w - \int \frac{1}{w + 1} d w

\int \frac{1}{w} d w = ln ∣ w ∣

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

= ln ∣ w ∣ - ln ∣ w + 1 ∣

Substitute back

= ln ∣ 9 e^{v} - 1 ∣ - ln ∣ 9 e^{v} - 1 + 1 ∣

Simplify

ln ∣ 9 e^{v} - 1 ∣ - ln ∣ 9 e^{v} - 1 + 1 ∣ : ln ∣ 9 e^{v} - 1 ∣ - 2 ln (3) - v

= ln ∣ 9 e^{v} - 1 ∣ - 2 ln (3) - v

Add a constant to the solution

= ln ∣ 9 e^{v} - 1 ∣ - 2 ln (3) - v + C

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What is the integral of 1/(9e^v-1) ?
The integral of 1/(9e^v-1) is ln|9e^v-1|-2ln(3)-v+C