What is the integral of (4^x)/(2^x-1) ?

The integral of (4^x)/(2^x-1) is 1/(ln(2))(2^x-1+ln|2^x-1|)+C

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integral of (4^x)/(2^x-1)

\int \frac{4 ^{x}}{2 ^{x} - 1} d x

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

Solution steps

\int \frac{4 ^{x}}{2 ^{x} - 1} d x

Apply u-substitution

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

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

Expand

\frac{u + 1}{u} : 1 + \frac{1}{u}

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

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{1}{ln ( 2 )} (u + ln ∣ u ∣)

Substitute back

u = 2^{x} - 1

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

Add a constant to the solution

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

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What is the integral of (4^x)/(2^x-1) ?
The integral of (4^x)/(2^x-1) is 1/(ln(2))(2^x-1+ln|2^x-1|)+C