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

The integral of 1/(sqrt(2^x)-1) is -x+2/(ln(2))ln|2^{x/2}-1|+C

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integral of 1/(sqrt(2^x)-1)

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{ln ( 2 ) u ( u - 1 )} d u

= \int \frac{1}{ln ( 2 ) 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}{ln ( 2 )} \cdot \int \frac{1}{u ( u - 1 )} d u

Apply u-substitution

: \int \frac{2}{v ( v - 1 )} d v

= \frac{1}{ln ( 2 )} \cdot \int \frac{2}{v ( v - 1 )} d v

Take the constant out:

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

= \frac{1}{ln ( 2 )} \cdot 2 \cdot \int \frac{1}{v ( v - 1 )} d v

Take the partial fraction of

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

= \frac{1}{ln ( 2 )} \cdot 2 \cdot \int - \frac{1}{v} + \frac{1}{v - 1} d v

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 )} \cdot 2 (- \int \frac{1}{v} d v + \int \frac{1}{v - 1} d v)

\int \frac{1}{v} d v = ln ∣ v ∣

\int \frac{1}{v - 1} d v = ln ∣ v - 1 ∣

= \frac{1}{ln ( 2 )} \cdot 2 (- ln ∣ v ∣ + ln ∣ v - 1 ∣)

Substitute back

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

Simplify

\frac{1}{ln ( 2 )} \cdot 2 (- ln 2^{x} + ln 2^{x} - 1) : - x + \frac{2}{ln ( 2 )} ln 2^{\frac{x}{2}} - 1

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

Add a constant to the solution

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

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What is the integral of 1/(sqrt(2^x)-1) ?
The integral of 1/(sqrt(2^x)-1) is -x+2/(ln(2))ln|2^{x/2}-1|+C