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

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

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

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

e Ei (- (1 + x)) + C

Solution steps

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

Apply u-substitution

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

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

Simplify

e^{- u + 1} : e^{- u} e

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= e \cdot \int \frac{e ^{- u}}{u} d u

Apply u-substitution

: \int \frac{e ^{v}}{v} d v

= e \cdot \int \frac{e ^{v}}{v} d v

Use the nonelementary integral:

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

= e Ei (v)

Substitute back

= e Ei (- (1 + x))

Add a constant to the solution

= e Ei (- (1 + x)) + C

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