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

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

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

- ln 1 + e^{x - 1} + x - 1 + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{1 + e ^{u}} d u

= \int \frac{1}{1 + e ^{u}} d u

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

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

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

\int 1 d u = u

= - ln ∣ 1 + e^{u} ∣ + u

Substitute back

u = x - 1

= - ln 1 + e^{x - 1} + x - 1

Add a constant to the solution

= - ln 1 + e^{x - 1} + x - 1 + C

\int 4 x - 3 sec (x) tan (x) d x

\int 4 cos^{2} (9 x) d x

\int \frac{( x ^{2} + 2 x + 4 )}{( x - 3 ) ^{5}} d x

\int \frac{1}{sin ^{2} ( x )} d sin (x) d x

\int x^{2} + e^{- x} d x

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