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

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

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

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

ln ∣ x ∣ + Ei (- x) + C

Solution steps

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

Expand

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

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

\int \frac{1}{x} d x = ln ∣ x ∣

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

= ln ∣ x ∣ + Ei (- x)

Add a constant to the solution

= ln ∣ x ∣ + Ei (- x) + C

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