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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

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

Add a constant to the solution

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

y = - 2 x - 1

n = 1 \sum \infty \frac{1}{n + 3 ^{n}}

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

f (x) = 9 x^{8} e^{x} + e^{x} x^{9}

\frac{d}{d x} (\frac{5 + csc ( x )}{9 - csc ( x )})

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