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

integral

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

\frac{e ^{2 x}}{x}

\frac{1}{( x + 1 ) ^{5}}

e^{1 + x}

\frac{1}{( x ( x + 1 ) ^{2} )}

f (x) = 2 x cos (x)

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