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

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

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integral of 1/((e^x-1))

integral

\frac{1}{( e ^{x} - 1 )}

ln ∣ e^{x} - 1 ∣ - x + C

Solution steps

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

\frac{1}{e ^{x} - 1} = \frac{e ^{x}}{e ^{x} - 1} - 1

= \int \frac{e ^{x}}{e ^{x} - 1} - 1 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{e ^{x}}{e ^{x} - 1} d x - \int 1 d x

\int \frac{e ^{x}}{e ^{x} - 1} d x = ln ∣ e^{x} - 1 ∣

\int 1 d x = x

= ln ∣ e^{x} - 1 ∣ - x

Add a constant to the solution

= ln ∣ e^{x} - 1 ∣ - x + C

in t e g r a l \frac{1}{( x ^{2} ( x + 1 ) ^{2} )}

in t e g r a l \frac{1}{( x ( x + 2 ) )}

in t e g r a l 5^{- x}

in t e g r a l \frac{y}{( x ^{2} + y ^{2} )}

in t e g r a l \frac{x ^{2}}{( ( x - 2 ) ( x ^{2} + 1 ) )}

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