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

The integral of e^x*(1-e^x)^{-1} is -ln|1-e^x|+C

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integral of e^x*(1-e^x)^{-1}

\int e^{x} \cdot (1 - e^{x})^{- 1} d x

- ln ∣ 1 - e^{x} ∣ + C

Solution steps

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

Simplify

e^{x} (1 - e^{x})^{- 1} : \frac{e ^{x}}{1 - e ^{x}}

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

Apply u-substitution

: \int - \frac{1}{u} d u

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - ln ∣ u ∣

Substitute back

u = 1 - e^{x}

= - ln ∣ 1 - e^{x} ∣

Add a constant to the solution

= - ln ∣ 1 - e^{x} ∣ + C

\frac{d}{d x} (2 x^{3} e^{x})

\frac{3}{x}

f (x) = x^{3} - 2 x + 1

- 3 sin (2 t)

\int x^{\frac{3}{2}} + 2 x + 1 d x

What is the integral of e^x*(1-e^x)^{-1} ?
The integral of e^x*(1-e^x)^{-1} is -ln|1-e^x|+C