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

The integral of sqrt(1-e^x) is -ln|sqrt(1-e^x)+1|+ln|sqrt(1-e^x)-1|+2sqrt(1-e^x)+C

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integral of sqrt(1-e^x)

\int 1 - e^{x} d x

- ln 1 - e^{x} + 1 + ln 1 - e^{x} - 1 + 2 1 - e^{x} + C

Solution steps

\int 1 - e^{x} d x

Apply u-substitution

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

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

Apply u-substitution

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

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

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int 1 d v = v

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

Substitute back

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

Simplify

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

= - ln 1 - e^{x} + 1 + ln 1 - e^{x} - 1 + 2 1 - e^{x}

Add a constant to the solution

= - ln 1 - e^{x} + 1 + ln 1 - e^{x} - 1 + 2 1 - e^{x} + C

x \to - \infty lim (x^{3} - x)

x \to \infty lim (x^{- 6})

\int - x \cdot e^{3 x} d x

\int x^{2} - y^{2} d y

\int \frac{4 x ^{3} - 6 x ^{4}}{2 x ^{6}} d x

What is the integral of sqrt(1-e^x) ?
The integral of sqrt(1-e^x) is -ln|sqrt(1-e^x)+1|+ln|sqrt(1-e^x)-1|+2sqrt(1-e^x)+C