What is the integral of (e^{2x})/(sqrt(1-e^x)) ?

The integral of (e^{2x})/(sqrt(1-e^x)) is -2(-1/3 (1-e^x)^{3/2}+sqrt(1-e^x))+C

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integral of (e^{2x})/(sqrt(1-e^x))

\int \frac{e ^{2 x}}{1 - e ^{x}} d x

- 2 (- \frac{1}{3} (1 - e^{x})^{\frac{3}{2}} + 1 - e^{x}) + C

Solution steps

\int \frac{e ^{2 x}}{1 - e ^{x}} d x

Apply u-substitution

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

Apply u-substitution

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

\int v^{2} d v = \frac{v ^{3}}{3}

\int 1 d v = v

= - 2 (- \frac{v ^{3}}{3} + v)

Substitute back

= - 2 (- \frac{( 1 - e ^{x} ) ^{3}}{3} + 1 - e^{x})

Simplify

- 2 (- \frac{( 1 - e ^{x} ) ^{3}}{3} + 1 - e^{x}) : - 2 (- \frac{1}{3} (1 - e^{x})^{\frac{3}{2}} + 1 - e^{x})

= - 2 (- \frac{1}{3} (1 - e^{x})^{\frac{3}{2}} + 1 - e^{x})

Add a constant to the solution

= - 2 (- \frac{1}{3} (1 - e^{x})^{\frac{3}{2}} + 1 - e^{x}) + C

What is the integral of (e^{2x})/(sqrt(1-e^x)) ?
The integral of (e^{2x})/(sqrt(1-e^x)) is -2(-1/3 (1-e^x)^{3/2}+sqrt(1-e^x))+C