What is the integral of (e^x)/(x+2) ?

The integral of (e^x)/(x+2) is 1/(e^2)\Ei(-(-x-2))+C

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integral of (e^x)/(x+2)

\int \frac{e ^{x}}{x + 2} d x

\frac{1}{e ^{2}} Ei (- (- x - 2)) + C

Solution steps

\int \frac{e ^{x}}{x + 2} d x

Factor

x + 2 : - (- x - 2)

= \int \frac{e ^{x}}{- ( - x - 2 )} d x

Take the constant out:

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

= - \int \frac{e ^{x}}{- x - 2} d x

Apply u-substitution

= - \int - \frac{e ^{- u - 2}}{u} d u

Take the constant out:

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

= - (- \int \frac{e ^{- u - 2}}{u} d u)

Simplify

e^{- u - 2} : e^{- u} e^{- 2}

= - (- \int \frac{e ^{- u} e ^{- 2}}{u} d u)

Take the constant out:

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

= - (- e^{- 2} \cdot \int \frac{e ^{- u}}{u} d u)

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= - (- \frac{1}{e ^{2}} \cdot \int \frac{e ^{- u}}{u} d u)

Apply u-substitution

= - (- \frac{1}{e ^{2}} \cdot \int \frac{e ^{v}}{v} d v)

Use the nonelementary integral:

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

= - (- \frac{1}{e ^{2}} Ei (v))

Substitute back

= - (- \frac{1}{e ^{2}} Ei (- (- x - 2)))

Simplify

= \frac{1}{e ^{2}} Ei (- (- x - 2))

Add a constant to the solution

= \frac{1}{e ^{2}} Ei (- (- x - 2)) + C

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What is the integral of (e^x)/(x+2) ?
The integral of (e^x)/(x+2) is 1/(e^2)\Ei(-(-x-2))+C