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

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

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integral of 1/(3^x-1)

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Take the partial fraction of

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

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

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

\int \frac{1}{u} d u = ln ∣ u ∣

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

Substitute back

u = 3^{x} - 1

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

Simplify

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

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

Add a constant to the solution

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

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