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

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

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

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

e^{x} + 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{e ^{- x}}{e ^{- 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{e ^{- x}}{e ^{- x}} d x

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

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

= e^{x} + x

Add a constant to the solution

= e^{x} + 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 e^x+x+C