What is the integral of 1/((2^t-1)) ?

The integral of 1/((2^t-1)) is (-ln(2)t+ln|-2^t+1|)/(ln(2))+C

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integral of 1/((2^t-1))

integral

\frac{1}{( 2 ^{t} - 1 )}

\frac{- ln ( 2 ) t + ln ∣ - 2 ^{t} + 1 ∣}{ln ( 2 )} + C

Solution steps

\int \frac{1}{2 ^{t} - 1} d t

Factor

2^{t} - 1 : - (- 2^{t} + 1)

= \int \frac{1}{- ( - 2 ^{t} + 1 )} d t

Take the constant out:

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

= - \int \frac{1}{- 2 ^{t} + 1} d t

Apply u-substitution

= - \int - \frac{1}{ln ( 2 ) ( - 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 ( 2 )} \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 ( 2 )} \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 ( 2 )} (- \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 ( 2 )} (- ln ∣ u - 1 ∣ + ln ∣ u ∣))

Substitute back

u = - 2^{t} + 1

= - (- \frac{1}{ln ( 2 )} (- ln - 2^{t} + 1 - 1 + ln - 2^{t} + 1))

Simplify

- (- \frac{1}{ln ( 2 )} (- ln - 2^{t} + 1 - 1 + ln - 2^{t} + 1)) : \frac{- ln ( 2 ) t + ln ∣ - 2 ^{t} + 1 ∣}{ln ( 2 )}

= \frac{- ln ( 2 ) t + ln ∣ - 2 ^{t} + 1 ∣}{ln ( 2 )}

Add a constant to the solution

= \frac{- ln ( 2 ) t + ln ∣ - 2 ^{t} + 1 ∣}{ln ( 2 )} + C

What is the integral of 1/((2^t-1)) ?
The integral of 1/((2^t-1)) is (-ln(2)t+ln|-2^t+1|)/(ln(2))+C