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

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

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

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

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

Solution steps

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

Factor

x - 1 : - (- x + 1)

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

Apply u-substitution

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

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

Simplify

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

Add a constant to the solution

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

x \to 0 lim (ln (1 - 5 x))

x y^{^{''}} - y^{^{'}} + y^{^{'}} = 0

\frac{d}{d x} (\frac{1 + x - 4 x}{x})

y = ln (3 x^{5})

\frac{d y}{d x} + 2 y = 2 e^{x}

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