What is the integral of (e^{-x})/(1-e^{-2x)} ?

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

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integral of (e^{-x})/(1-e^{-2x)}

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

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

Solution steps

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

Take the partial fraction of

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

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (x^{2} - y^{2} + x + 1)

\frac{2 x ^{2} + 4 x + 6}{x}

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

- 4 \frac{d y}{d x} + 12 x^{2} = 0

y = 2 x - 3

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