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

The integral of (1+e^{-2x})/(e^x) is -1/(e^x)-1/3 e^{-3x}+C

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

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

- \frac{1}{e ^{x}} - \frac{1}{3} e^{- 3 x} + C

Solution steps

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

Expand

\frac{1 + e ^{- 2 x}}{e ^{x}} : \frac{1}{e ^{x}} + \frac{1}{e ^{3 x}}

= \int \frac{1}{e ^{x}} + \frac{1}{e ^{3 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 ^{3 x}} d x

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

\int \frac{1}{e ^{3 x}} d x = - \frac{1}{3} e^{- 3 x}

= - \frac{1}{e ^{x}} - \frac{1}{3} e^{- 3 x}

Add a constant to the solution

= - \frac{1}{e ^{x}} - \frac{1}{3} e^{- 3 x} + C

\int_{0}^{3} 3 x^{2} d x

\frac{\partial}{\partial x} (x^{\frac{2}{5}} \cdot y^{\frac{3}{5}})

\int \frac{( 3 x + 5 )}{x ^{2} + x + 1} d x

\frac{d}{d x} (\frac{1}{5 + 2 x})

y = 5 x - 2, (5, 23)

What is the integral of (1+e^{-2x})/(e^x) ?
The integral of (1+e^{-2x})/(e^x) is -1/(e^x)-1/3 e^{-3x}+C