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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

= - \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

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

\frac{p}{( p ^{2} + 4 )}

4 x^{2} + 3 x + 2 \cdot 2 (x + 1)

(8 x - x^{4})^{3} (8 x^{3} - 16)

cos^{2} (\frac{x}{4})

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