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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

integral

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

\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

= - \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

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

= - (- \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

e^{x} cos^{2} (x)

y = cos (x^{2}) (0.1)

t^{2} (e^{t})^{2}

4 x^{10}

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

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