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

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

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

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

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

Solution steps

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

Expand

\frac{1 - e ^{- x}}{e ^{x}} : \frac{1}{e ^{x}} - \frac{1}{e ^{2 x}}

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

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

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

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

Simplify

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

Add a constant to the solution

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

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