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

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

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integral of 2/(1-e^x)

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

2 (- ln ∣ 1 - e^{x} ∣ + x) + C

Solution steps

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

Take the constant out:

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

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

\frac{1}{1 - e ^{x}} = \frac{e ^{x}}{1 - e ^{x}} + 1

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

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

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

\int 1 d x = x

= 2 (- ln ∣ 1 - e^{x} ∣ + x)

Add a constant to the solution

= 2 (- ln ∣ 1 - e^{x} ∣ + x) + C

n = 1 \sum \infty \frac{2 n}{9 n - 1}

\frac{\partial}{\partial x} (x y (x - y)^{- 1})

x \to 99 lim (5)

\int_{0}^{\infty} x e^{- 3 x} d x

x \to 0 + lim (csc (x) - \frac{1}{x})

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