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

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

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integral of 1/(1+e^x)

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

- ln ∣ 1 + e^{x} ∣ + x + C

Solution steps

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

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

= \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

= - \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

= - ln ∣ 1 + e^{x} ∣ + x

Add a constant to the solution

= - ln ∣ 1 + e^{x} ∣ + x + C

- x^{2} + 10 x - 16, 2 x - 4

\int \frac{1}{x ^{\frac{7}{2}}} d x

\frac{d}{d x} ((f (x) (4 x))^{2} + 5)

6 x + y^{2} = 16, x = y

\frac{d}{d x} (3 a x^{4} - 4 a x^{3} - 5 b x^{2} + 7 c x)