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

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

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integral of (e^x)/((x+1))

integral

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

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

Solution steps

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

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

Take the constant out:

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

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

Apply exponent rule:

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

= \frac{1}{e} \cdot \int \frac{e ^{u}}{u} d u

Use the nonelementary integral:

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

= \frac{1}{e} Ei (u)

Substitute back

u = x + 1

= \frac{1}{e} Ei (x + 1)

Add a constant to the solution

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

\frac{1}{1 + t ^{2}}

\frac{( x )}{( 1 + x ^{\frac{1}{4}} )}

\frac{x ^{2}}{9}

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

\frac{6}{( 1 + t ^{2} )}

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