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

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

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

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

x + e^{x} + C

Solution steps

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

Expand

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

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

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

= x + e^{x}

Add a constant to the solution

= x + e^{x} + C

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