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

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

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integral of 1/(e^x+1)

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

- ln ∣ e^{x} + 1 ∣ + x + C

Solution steps

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

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

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

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

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

\int 1 d x = x

= - ln ∣ e^{x} + 1 ∣ + x

Add a constant to the solution

= - ln ∣ e^{x} + 1 ∣ + x + C

\frac{\partial}{\partial t} (s \cdot cos (t))

\frac{d}{d y} (1 + y^{2})

f (x) = 2 x^{2} + 4, at x = - 2

z \to 1 + i lim (z^{2} - 5 z + 10)

y^{^{'}} = 2 y - 4 x