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

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

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

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

x - 2 ln ∣ e^{x} + 1 ∣ + C

Solution steps

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

Apply u-substitution

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

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

Take the partial fraction of

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

f (x) = x sin (x)

y^{^{'}} = e^{3 x + 2 y}

e^{2 x - 3}

\frac{\partial}{\partial x} (\frac{1}{3 x ^{2} + y})

x \to - 3 lim (g (x) (f (x)))

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