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

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

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

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

1 - e^{x} - ln ∣ 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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Expand

\frac{- v + 1}{v} : - 1 + \frac{1}{v}

= - \int - 1 + \frac{1}{v} 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

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

\int 1 d v = v

\int \frac{1}{v} d v = ln ∣ v ∣

= - (- v + ln ∣ v ∣)

Substitute back

= - (- (1 - e^{x}) + ln ∣ 1 - e^{x} ∣)

Simplify

= 1 - e^{x} - ln ∣ 1 - e^{x} ∣

Add a constant to the solution

= 1 - e^{x} - ln ∣ 1 - e^{x} ∣ + C

\frac{d}{d x} (ln (\frac{x + 2}{x - 1}))

y = \frac{x + 2 ^{x}}{cos ( π x )}

ln (arctan (x))

\frac{d y}{d x} = 2 + \frac{1}{x}, y (1) = 5

y^{^{'}} + 2 x y = 2 x^{3}, y (0) = 2

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