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

The integral of 1/(sqrt(1-e^x)) is -ln|sqrt(1-e^x)+1|+ln|sqrt(1-e^x)-1|+C

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

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

- ln 1 - e^{x} + 1 + ln 1 - e^{x} - 1 + C

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

: \int - \frac{2}{- v ^{2} + 1} d v

= \int - \frac{2}{- v ^{2} + 1} d v

Take the constant out:

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

= - 2 \cdot \int \frac{1}{- v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= - 2 (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

Substitute back

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

Simplify

- 2 (\frac{ln 1 - e ^{x} + 1}{2} - \frac{ln 1 - e ^{x} - 1}{2}) : - ln 1 - e^{x} + 1 + ln 1 - e^{x} - 1

= - ln 1 - e^{x} + 1 + ln 1 - e^{x} - 1

Add a constant to the solution

= - ln 1 - e^{x} + 1 + ln 1 - e^{x} - 1 + C

\frac{\partial}{\partial x} (x^{4} y^{3} - y^{4})

\frac{d}{d y} (ln (x + 2 y + 3 z))

x \to 5 - lim (\frac{2}{x - 5} + 1)

- 4 x^{2} + 1

\frac{\partial}{\partial y} (sin (x^{5} - 7 y))

What is the integral of 1/(sqrt(1-e^x)) ?
The integral of 1/(sqrt(1-e^x)) is -ln|sqrt(1-e^x)+1|+ln|sqrt(1-e^x)-1|+C