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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(1+e^z)

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

- ln ∣ 1 + e^{z} ∣ + z + C

Solution steps

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

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

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

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 ^{z}}{1 + e ^{z}} d z + \int 1 d z

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

\int 1 d z = z

= - ln ∣ 1 + e^{z} ∣ + z

Add a constant to the solution

= - ln ∣ 1 + e^{z} ∣ + z + C

x + y^{2} = 6, x + y = 0

y = (x^{6} - 7 x + 4) (x^{4} - 5 x)

f (x) = x^{2} + 4

(\frac{y}{x} + 3 x) d x + (ln (x) - 2) d y = 0

\int \frac{x ^{2} - 2 x}{x ^{3} - 3 x ^{2}} d x