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

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

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

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

x - 2 ln ∣ e^{x} + 1 ∣ + C

Solution steps

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

Apply u-substitution

: \int \frac{1 - u}{u ( u + 1 )} d u

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

Take the partial fraction of

\frac{1 - u}{u ( u + 1 )} : \frac{1}{u} - \frac{2}{u + 1}

= \int \frac{1}{u} - \frac{2}{u + 1} d u

Apply the Sum Rule:

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

= \int \frac{1}{u} d u - \int \frac{2}{u + 1} d u

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{2}{u + 1} d u = 2 ln ∣ u + 1 ∣

= ln ∣ u ∣ - 2 ln ∣ u + 1 ∣

Substitute back

u = e^{x}

= ln ∣ e^{x} ∣ - 2 ln ∣ e^{x} + 1 ∣

ln ∣ e^{x} ∣ = x

= x - 2 ln ∣ e^{x} + 1 ∣

Add a constant to the solution

= x - 2 ln ∣ e^{x} + 1 ∣ + C

g (x) = (\frac{9 + 6 x}{7 - 4 x})

\frac{\partial}{\partial z} (\frac{y}{z})

\frac{x}{1 + x}

\frac{d}{d x} ((A x^{2} + B x) e^{- x})

\frac{d}{d t} (\frac{2 t}{t ^{2} + 2})

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